From 02b56c937fa259c22238d1e44d84110120ff85fe Mon Sep 17 00:00:00 2001 From: Leni Aniva Date: Sat, 5 Oct 2024 01:01:27 -0700 Subject: [PATCH] chore: Remove duplicate files --- .../dsp_debug2_sf/n_plus_zero_full_proof.json | 10 - .../zero_plus_n_informal_full_proof.json | 6 - .../zero_plus_n_informal_proof.json | 6 - .../debug/toy_example1_dsp/dsp_debug3_sf.json | 25 -- .../debug/toy_example1_dsp/dsp_debug4_sf.json | 368 ------------------ .../dsp_debug5_sf/dsp_debug5_sf_test.json | 98 ----- .../dsp_debug5_sf/dsp_debug5_sf_train.json | 117 ------ .../toy_example1_dsp/dsp_debug5_sf/readme.md | 106 ----- 8 files changed, 736 deletions(-) delete mode 100644 data/debug/toy_example1_dsp/dsp_debug2_sf/n_plus_zero_full_proof.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_full_proof.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_proof.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug3_sf.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug4_sf.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_test.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_train.json delete mode 100644 data/debug/toy_example1_dsp/dsp_debug5_sf/readme.md diff --git a/data/debug/toy_example1_dsp/dsp_debug2_sf/n_plus_zero_full_proof.json b/data/debug/toy_example1_dsp/dsp_debug2_sf/n_plus_zero_full_proof.json deleted file mode 100644 index 4c68038..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug2_sf/n_plus_zero_full_proof.json +++ /dev/null @@ -1,10 +0,0 @@ -{ - "problem": "For any natural number n, n + 0 = n.", - "level": "SF foundations level 1", - "type": "Algebra", - "solution": [ - "Consider some natural number n. The proof will be by induction. Consider the base case n=0. We have 0 + 0 = 0 which holds by the definition of addion.", - "Now the inductive case we want to show n'+1 + 0 = n'+1. Assume it holds for n i.e., n' + 0 = n. So now by the we know the LHS is equal to (n'+0) + 1.", - "Then by the induction hypothesis we know the LHS simplifies to n'+1 completing the proof. Qed." - ] -} \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_full_proof.json b/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_full_proof.json deleted file mode 100644 index c7f9bea..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_full_proof.json +++ /dev/null @@ -1,6 +0,0 @@ -{ - "problem": "For any natural number n, 0 + n = n.", - "level": "SF foundations level 1", - "type": "Algebra", - "solution": "Consider some natural number n. We now want to show 0 + n = n. Using addition both sides are equal." -} \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_proof.json b/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_proof.json deleted file mode 100644 index 952f3d2..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug2_sf/zero_plus_n_informal_proof.json +++ /dev/null @@ -1,6 +0,0 @@ -{ - "problem": "For any natural number n, 0 + n = n.", - "level": "SF foundations level 1", - "type": "Algebra", - "solution": "Consider some natural number n. We now want to show 0 + n = n. For that we use the definition of addition to get n = n and conclude both sides are equal (by reflexity)." -} \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug3_sf.json b/data/debug/toy_example1_dsp/dsp_debug3_sf.json deleted file mode 100644 index eb5f978..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug3_sf.json +++ /dev/null @@ -1,25 +0,0 @@ -[ - { - "problem": "For any natural number n, 0 + n = n.", - "level": "SF foundations level 1", - "type": "Logical Foundations", - "solution": [ - "Consider some natural number n. We want to show 0 + n = n. ", - "By using definition of addition on both sides, LHS and RHS are now equal, done." - ] - }, - { - "problem": "For any natural number n, n + 0 = n.", - "level": "SF foundations level 1", - "type": "Logical Foundations", - "solution": [ - "Consider some natural number n. The proof will be by induction. ", - "The base case n=0, so we have 0 + 0 = 0, which holds by the definition of addion. ", - "Consider the inductive case, so we want to show (k + 1) + 0 = (k + 1) for any k < n assuming the IH holds for such k (IH: k + 0 = k). ", - "By the IH we have (k + 1) + 0 = (k + 1). ", - "By def of addition we have (k + 0) + 1 = (k + 1). ", - "By the induction hypothesis (IH) we have k + 0 = k. ", - "LHS and RHS are equal so proof is complete." - ] - } -] \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug4_sf.json b/data/debug/toy_example1_dsp/dsp_debug4_sf.json deleted file mode 100644 index aae5f3a..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug4_sf.json +++ /dev/null @@ -1,368 +0,0 @@ -[ - { - "nl_problem": "For any natural number n, n + 0 = n.", - "nl_solution": [ - "Consider some natural number n. We want to show n + 0 = n. ", - "By using fats of addition on both sides, LHS and RHS are now equal, done." - ], - "fl_problem": "theorem n_plus_zero : ∀ n : ℕ, n + 0 = n := by", - "fl_solution": [ - "-- Prove that n + 0 = n.\n", - "theorem n_plus_zero : ∀ n : ℕ, n + 0 = n := by\n", - " -- Consider some n in Nats.\n", - " intro n", - "-- Using facts of addition simplify n + 0 to n.\n", - " rw [Nat.add_zero]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ], - - "nl_problem_proved_sketch": "For any natural number n, n + 0 = n.", - "nl_solution_proved_sketch": [ - "We want to show n + 0 = n. ", - "We have the fact of addition that, n + 0 = n. ", - "Thus, the left-hand side and right-hand side are equal, which completes the proof." - ], - "fl_problem_proved_sketch": "theorem n_plus_zero_proved_formal_sketch : ∀ n : ℕ, n + 0 = n := by", - "fl_solution_proved_sketch": [ - "-- Prove that n + 0 = n via a formal proof sketch", - "theorem n_plus_zero_proved_formal_sketch : ∀ n : ℕ, n + 0 = n := by", - " -- We have the fact of addition n + 0 = n, use it to show left and right are equal.", - " have h_nat_add_zero: ∀ n : ℕ, n + 0 = n := Nat.add_zero", - " exact h_nat_add_zero" - ], - "src_header_fl_solution_proved_sketch": ["import Mathlib.Data.Nat.Basic"], - - "nl_problem_proved_sketch_aesop": "For any natural number n, n + 0 = n.", - "nl_solution_proved_sketch_aesop": [ - "We want to show n + 0 = n. ", - "We have the fact of addition that, n + 0 = n. ", - "Thus, the left-hand side and right-hand side are equal, which completes the proof." - ], - "fl_problem_proved_sketch_aesop": "theorem n_plus_zero_proved_formal_sketch' : ∀ n : ℕ, n + 0 = n := by", - "fl_solution_proved_sketch_aesop": [ - "-- Prove that n + 0 = n via a formal proof sketch with aesop. ", - "theorem n_plus_zero_proved_formal_sketch' : ∀ n : ℕ, n + 0 = n := by", - " -- We have the fact of addition n + 0 = n, use it to show left and right are equal. ", - " have h_nat_add_zero: ∀ n : ℕ, n + 0 = n := by aesop", - " exact h_nat_add_zero" - ], - "src_header_fl_solution_proved_sketch_aesop": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ] - }, - - - - { - "nl_problem": "For any natural number n, 0 + n = n.", - "nl_solution": [ - "Consider some natural number n. We want to show 0 + n = n.", - "By using facts of addition and induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem zero_plus_n : ∀ n : ℕ, 0 + n = n := by", - "fl_solution": [ - "-- Prove that 0 + n = n by induction", - "theorem zero_plus_n : ∀ n : ℕ, 0 + n = n := by", - "-- Consider some n in Nats.", - "intro n", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: 0 + 0 = 0", - "rw [Nat.add_zero]", - "| succ n ih =>", - "-- Inductive step: assume 0 + n = n, prove 0 + succ n = succ n", - "rw [Nat.add_succ]", - "rw [ih]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ], - - "nl_problem_proved_sketch": "For any natural number n, 0 + n = n.", - "nl_solution_proved_sketch": [ - "We want to show 0 + n = n.", - "By using the fact of addition and performing induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem_proved_sketch": "theorem zero_plus_n_proved_formal_sketch : ∀ n : ℕ, 0 + n = n := by", - "fl_solution_proved_sketch": [ - "-- Prove that 0 + n = n by induction via a formal proof sketch", - "theorem zero_plus_n_proved_formal_sketch : ∀ n : ℕ, 0 + n = n := by", - "-- Consider some n in Nats.", - "intro n", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: 0 + 0 = 0", - "have h_base: 0 + 0 = 0 := by rw [Nat.add_zero]", - "exact h_base", - "| succ n ih =>", - "-- Inductive step: assume 0 + n = n, prove 0 + succ n = succ n", - "have h_inductive: 0 + Nat.succ n = Nat.succ n := by", - "rw [Nat.add_succ]", - "rw [ih]", - "exact h_inductive" - ], - "src_header_fl_solution_proved_sketch": [ - "import Mathlib.Data.Nat.Basic" - ], - - "nl_problem_proved_sketch_aesop": "For any natural number n, 0 + n = n.", - "nl_solution_proved_sketch_aesop": [ - "We want to show 0 + n = n.", - "By using the fact of addition and performing induction on n, we can prove the statement for both the base case and the inductive step using aesop." - ], - - "fl_problem_proved_sketch_aesop": "theorem zero_plus_n_proved_formal_sketch' : ∀ n : ℕ, 0 + n = n := by", - "fl_solution_proved_sketch_aesop": [ - "-- Prove that 0 + n = n by induction via a formal proof sketch with aesop.", - "theorem zero_plus_n_proved_formal_sketch' : ∀ n : ℕ, 0 + n = n := by", - "-- Consider some n in Nats.", - "intro n", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: 0 + 0 = 0", - "have h_base: 0 + 0 = 0 := by aesop", - "exact h_base", - "| succ n ih =>", - "-- Inductive step: assume 0 + n = n, prove 0 + succ n = succ n", - "have h_inductive: 0 + Nat.succ n = Nat.succ n := by aesop", - "exact h_inductive" - ], - "src_header_fl_solution_proved_sketch_aesop": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ] - }, - - - - { - "nl_problem": "For any natural numbers n and m we have commutativity, n + m = m + n.", - "nl_solution": [ - "Consider some natural numbers n and m. We want to show n + m = m + n.", - "By using facts of addition and induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_comm_normal : ∀ n m : ℕ, n + m = m + n := by", - "fl_solution": [ - "-- Prove that n + m = m + n", - "theorem add_comm_normal : ∀ n m : ℕ, n + m = m + n := by", - "-- Consider some n and m in Nats.", - "intros n m", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + m = m + 0.", - "-- Using the definition of addition, 0 + m = m and m + 0 = m.", - "rw [Nat.zero_add, Nat.add_zero]", - "| succ n ih =>", - "-- Inductive step: Assume n + m = m + n, we need to show succ n + m = m + succ n.", - "-- We use the fact n + (m + 1) = (n + m) + 1.", - "have plus_n_Sm_normal: ∀ n m : ℕ, n + (m + 1) = (n + m) + 1 := by", - " intros n m", - " rw [Nat.add_succ]", - "-- Apply the fact to rewrite succ n + m = (n + m) + 1.", - "rw [Nat.add_succ, Nat.add_zero]", - "rw [← ih]", - "rw [Nat.succ_add]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ] - }, - - { - "nl_problem": "For any natural numbers n and m, n + m = m + n.", - "nl_solution": [ - "Consider some natural numbers n and m. We want to show n + m = m + n.", - "By using the fact of addition and performing induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_comm_proved_formal_sketch : ∀ n m : ℕ, n + m = m + n := by", - "fl_solution": [ - "-- Prove that n + m = m + n via a formal proof sketch", - "theorem add_comm_proved_formal_sketch : ∀ n m : ℕ, n + m = m + n := by", - "-- Consider some n and m in Nats.", - "intros n m", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + m = m + 0.", - "-- We have the fact 0 + m = m by the definition of addition.", - "have h_base: 0 + m = m := Nat.zero_add m", - "-- We also have the fact m + 0 = m by the definition of addition.", - "have h_symm: m + 0 = m := Nat.add_zero m", - "-- Combining these, we get 0 + m = m + 0.", - "rw [h_base, h_symm]", - "| succ n ih =>", - "-- Inductive step: Assume n + m = m + n, we need to show succ n + m = m + succ n.", - "-- By the inductive hypothesis, we have n + m = m + n.", - "have h_inductive: n + m = m + n := ih", - "-- proof is:", - "-- We eventually want to flip n + m and simplify to make both sides the same. Thus,", - "-- 1. Note we start with: Nat.succ n + m = m + Nat.succ n, so, pull the succ out from m + Nat.succ n on the right side from the addition using addition facts Nat.add_succ.", - "have h_pull_succ_out_from_right: m + Nat.succ n = Nat.succ (m + n) := by rw [Nat.add_succ]", - "-- 2. then to flip m + S n to something like S (n + m) we need to use the IH.", - "have h_flip_n_plus_m: Nat.succ (n + m) = Nat.succ (m + n) := by rw [h_inductive]", - "-- 3. Now the n & m are on the correct sides Nat.succ n + m = Nat.succ (n + m), so let's use the def of addition to pull out the succ from the addition on the left using Nat.succ_add.", - "have h_pull_succ_out_from_left: Nat.succ n + m = Nat.succ (n + m) := by rw [Nat.succ_add]", - "-- Combining these, we get succ n + m = m + succ n.", - "rw [h_pull_succ_out_from_right, ←h_flip_n_plus_m, h_pull_succ_out_from_left]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ] - }, - - { - "nl_problem": "For any natural numbers n and m, n + m = m + n.", - "nl_solution": [ - "Consider some natural numbers n and m. We want to show n + m = m + n.", - "By using the fact of addition and performing induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_comm_proved_formal_sketch_aesop : ∀ n m : ℕ, n + m = m + n := by", - "fl_solution": [ - "-- Prove that n + m = m + n via a formal proof sketch with aesop.", - "theorem add_comm_proved_formal_sketch_aesop : ∀ n m : ℕ, n + m = m + n := by", - "-- Consider some n and m in Nats.", - "intros n m", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + m = m + 0.", - "-- We have the fact 0 + m = m by the definition of addition.", - "have h_base: 0 + m = m := by aesop", - "-- We also have the fact m + 0 = m by the definition of addition.", - "have h_symm: m + 0 = m := by aesop", - "-- Combining these, we get 0 + m = m + 0.", - "rw [h_base, h_symm]", - "| succ n ih =>", - "-- Inductive step: Assume n + m = m + n, we need to show succ n + m = m + succ n.", - "-- By the inductive hypothesis, we have n + m = m + n.", - "have h_inductive: n + m = m + n := by aesop", - "-- proof is:", - "-- We eventually want to flip n + m and simplify to make both sides the same. Thus,", - "-- 1. Note we start with: Nat.succ n + m = m + Nat.succ n, so, pull the succ out from m + Nat.succ n on the right side from the addition using addition facts Nat.add_succ.", - "have h_pull_succ_out_from_right: m + Nat.succ n = Nat.succ (m + n) := by aesop", - "-- 2. then to flip m + S n to something like S (n + m) we need to use the IH.", - "have h_flip_n_plus_m: Nat.succ (n + m) = Nat.succ (m + n) := by aesop", - "-- 3. Now the n & m are on the correct sides Nat.succ n + m = Nat.succ (n + m), so let's use the def of addition to pull out the succ from the addition on the left using Nat.succ_add.", - "have h_pull_succ_out_from_left: Nat.succ n + m = Nat.succ (n + m) := by rw [Nat.succ_add]", - "-- Combining these, we get succ n + m = m + succ n.", - "rw [h_pull_succ_out_from_right, ←h_flip_n_plus_m, h_pull_succ_out_from_left]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ] - }, - - - - { - "nl_problem": "Prove that for any natural numbers n, m, and p, n + (m + p) = (n + m) + p.", - "nl_solution": [ - "Consider some natural numbers n, m, and p. We want to show n + (m + p) = (n + m) + p.", - "By using facts of addition and induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_assoc_normal : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "fl_solution": [ - "-- Prove that n + (m + p) = (n + m) + p", - "theorem add_assoc_normal : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "-- Consider some n, m, and p in Nats.", - "intros n m p", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + (m + p) = (0 + m) + p.", - "-- Using the definition of addition, 0 + (m + p) = m + p and (0 + m) + p = m + p.", - "rw [Nat.zero_add, Nat.zero_add]", - "| succ n ih =>", - "-- Inductive step: Assume n + (m + p) = (n + m) + p, we need to show succ n + (m + p) = (succ n + m) + p.", - "-- proof strategy is, we move succ n out (or in) enough times then use the IH until both sides are the same.", - "-- 1. let's start by pulling out the succ from the left side and have the entire addition inside the succ.", - "rw [Nat.succ_add]", - "-- 2. Now that we have the IH hypothesis appearing inside the left, let's apply it so we have n + (m + p) = (n + m) + p.", - "rw [ih]", - "-- 3. Now that the parentheses (apps of plus) are in the right place for both sides, push the succ on the left twice so both terms are the same.", - "rw [← Nat.succ_add, ← Nat.succ_add]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ] - }, - - { - "nl_problem": "Prove that for any natural numbers n, m, and p, n + (m + p) = (n + m) + p.", - "nl_solution": [ - "Consider some natural numbers n, m, and p. We want to show n + (m + p) = (n + m) + p.", - "By using facts of addition and induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_assoc_proved_formal_sketch : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "fl_solution": [ - "-- Prove that n + (m + p) = (n + m) + p", - "theorem add_assoc_proved_formal_sketch : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "-- Consider some n, m, and p in Nats.", - "intros n m p", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + (m + p) = (0 + m) + p.", - "-- Using the definition of addition, 0 + (m + p) = m + p and (0 + m) + p = m + p.", - "rw [Nat.zero_add, Nat.zero_add]", - "| succ n ih =>", - "-- Inductive step: Assume n + (m + p) = (n + m) + p, we need to show succ n + (m + p) = (succ n + m) + p.", - "-- proof strategy is, we move succ n out (or in) enough times then use the IH until both sides are the same.", - "-- 1. let's start by pulling out the succ from the left side and have the entire addition inside the succ.", - "have h_pull_add_succ_out_from_left: Nat.succ n + (m + p) = Nat.succ (n + (m + p)) := by rw [Nat.succ_add]", - "-- 2. Now that we have the IH hypothesis appearing inside the left, let's apply it so we have n + (m + p) = (n + m) + p.", - "have h_inside_left_associates: Nat.succ (n + (m + p)) = Nat.succ ((n + m) + p) := by rw [ih]", - "-- 3. Now that the parentheses (apps of plus) are in the right place for both sides, push the succ on the left twice so both terms are the same.", - "have h_push_succ_in_left_twice: Nat.succ ((n + m) + p) = ((Nat.succ n) + m) + p := by rw [← Nat.succ_add, ← Nat.succ_add]", - "-- Combining these, we get succ n + (m + p) = (succ n + m) + p.", - "rw [h_pull_add_succ_out_from_left, h_inside_left_associates, h_push_succ_in_left_twice]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic" - ] - }, - - { - "nl_problem": "Prove that for any natural numbers n, m, and p, n + (m + p) = (n + m) + p.", - "nl_solution": [ - "Consider some natural numbers n, m, and p. We want to show n + (m + p) = (n + m) + p.", - "By using facts of addition and induction on n, we can prove the statement for both the base case and the inductive step." - ], - "fl_problem": "theorem add_assoc_proved_formal_sketch_aesop : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "fl_solution": [ - "-- Prove that n + (m + p) = (n + m) + p via a formal proof sketch with aesop", - "theorem add_assoc_proved_formal_sketch_aesop : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by", - "-- Consider some n, m, and p in Nats.", - "intros n m p", - "-- Perform induction on n.", - "induction n with", - "| zero =>", - "-- Base case: When n = 0, we need to show 0 + (m + p) = (0 + m) + p.", - "-- Using the definition of addition, 0 + (m + p) = m + p and (0 + m) + p = m + p.", - "rw [Nat.zero_add, Nat.zero_add]", - "| succ n ih =>", - "-- Inductive step: Assume n + (m + p) = (n + m) + p, we need to show succ n + (m + p) = (succ n + m) + p.", - "-- proof strategy is, we move succ n out (or in) enough times then use the IH until both sides are the same.", - "-- 1. let's start by pulling out the succ from the left side and have the entire addition inside the succ.", - "have h_pull_add_succ_out_from_left: Nat.succ n + (m + p) = Nat.succ (n + (m + p)) := by rw [Nat.succ_add]", - "-- 2. Now that we have the IH hypothesis appearing inside the left, let's apply it so we have n + (m + p) = (n + m) + p.", - "have h_inside_left_associates: Nat.succ (n + (m + p)) = Nat.succ ((n + m) + p) := by aesop", - "-- 3. Now that the parentheses (apps of plus) are in the right place for both sides, push the succ on the left twice so both terms are the same.", - "have h_push_succ_in_left_twice: Nat.succ ((n + m) + p) = ((Nat.succ n) + m) + p := by rw [← Nat.succ_add, ← Nat.succ_add]", - "-- Combining these, we get succ n + (m + p) = (succ n + m) + p.", - "rw [h_pull_add_succ_out_from_left, h_inside_left_associates, h_push_succ_in_left_twice]" - ], - "src_header_fl_solution": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ] - } -] \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_test.json b/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_test.json deleted file mode 100644 index 770ce32..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_test.json +++ /dev/null @@ -1,98 +0,0 @@ -[ - { - "nl_problem": ["Prove that for any natural number n, 0 + n = n."], - "nl_solution": [ - "Consider any natural number n. We will prove the statement by induction on n.", - "Base case: When n = 0, we need to show that 0 + 0 = 0. This is true by the definition of addition.", - "Inductive step: Assume that for some natural number n, 0 + n = n. We need to show that 0 + (n + 1) = (n + 1). By the definition of addition and the inductive hypothesis, we have 0 + (n + 1) = (0 + n) + 1 = n + 1. Therefore, the statement holds for n + 1.", - "Thus, by induction, we have proved that for any natural number n, 0 + n = n." - ], - "nl_solution_sketch": [ - "Consider any natural number n, and do induction on n.", - "Base case: 0 + 0 = 0 by properties of addition.", - "Inductive step we have 0 + n = n. Then 0 + (n + 1) = (0 + n) + 1 = n + 1.", - "Where, 0 + n = n by assumption,qed." - ], - "fl_problem": ["theorem zero_plus_n_proved_formal_sketch : ∀ n : ℕ, 0 + n = n := "], - "fl_partial_sketch": [ - "by\n", - " -- Consider some n in Nats.\n", - " intro n\n", - " -- Perform induction on n.\n", - " induction n with\n", - " | zero =>\n", - " -- Base case: 0 + 0 = 0\n", - " have h_base: 0 + 0 = 0 := \n", - " -- Combine facts to close goal\n", - " \n", - " | succ n ih =>\n", - " -- Inductive step: assume 0 + n = n, prove 0 + succ n = succ n\n", - " have h_inductive: 0 + Nat.succ n = Nat.succ n := \\n", - " -- Combine facts to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "1" - }, - - - - { - "nl_problem": ["Prove that for any natural number n, m, and p, n + (m + p) = (n + m) + p."], - "nl_solution": [ - "Consider any natural numbers n, m, and p. We will prove the statement by induction on n.", - "Base case: When n = 0, we need to show that 0 + (m + p) = (0 + m) + p. By the definition of addition, we have 0 + (m + p) = m + p and (0 + m) + p = m + p. Therefore, 0 + (m + p) = (0 + m) + p.", - "Inductive step: Assume that for some natural number n, n + (m + p) = (n + m) + p. We need to show that (n + 1) + (m + p) = ((n + 1) + m) + p.", - "1. First, pull out the successor from the left side to have the entire addition inside the successor: (n + 1) + (m + p) = (n + (m + p)) + 1.", - "2. By the inductive hypothesis, we know that n + (m + p) = (n + m) + p. So we can replace n + (m + p) with (n + m) + p inside the successor: (n + (m + p)) + 1 = ((n + m) + p) + 1.", - "3. Finally, push the successor on the left twice to align both sides: ((n + m) + p) + 1 = (n + 1) + (m + p) = ((n + 1) + m) + p.", - "Thus, by induction, we have proved that for any natural numbers n, m, and p, n + (m + p) = (n + m) + p." - ], - "nl_solution_sketch": [ - "Consider any natural numbers n, m, and p. We will do induction on n.", - "Base case: 0 + (m + p) = (0 + m) + p by properties of addition.", - "Inductive step, we have n + (m + p) = (n + m) + p. Then (n + 1) + (m + p) = (n + (m + p)) + 1 = ((n + m) + p) + 1.", - "Thus, (n + 1) + (m + p) = ((n + 1) + m) + p, qed." - ], - "fl_problem": ["theorem add_assoc_proved_formal_sketch : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := "], - "fl_partial_sketch": [ - "by\n", - " -- Consider some n, m, and p in Nats.\n", - " intros n m p\n", - " -- Perform induction on n.\n", - " induction n with\n", - " | zero =>\n", - " -- Base case: When n = 0, we need to show 0 + (m + p) = (0 + m) + p.\n", - " -- We have the fact 0 + (m + p) = m + p by the definition of addition.\n", - " have h_base: 0 + (m + p) = m + p := \n", - " -- We also have the fact (0 + m) + p = m + p by the definition of addition.\n", - " have h_symm: (0 + m) + p = m + p := \n", - " -- Combine facts to close goal\n", - " \n", - " | succ n ih =>\n", - " -- Inductive step: Assume n + (m + p) = (n + m) + p, we need to show succ n + (m + p) = (succ n + m) + p.\n", - " -- By the inductive hypothesis, we have n + (m + p) = (n + m) + p.\n", - " have h_inductive: n + (m + p) = (n + m) + p := \n", - " -- 1. Let's start by pulling out the succ from left side and have the entire addition inside the succ.\n", - " have h_pull_succ_out_from_left: Nat.succ n + (m + p) = Nat.succ (n + (m + p)) := \n", - " -- 2. Now that we have the IH hypothesis appearing inside the left, let's apply it so we have n + (m + p) = (n + m) + p.\n", - " have h_inside_left_associates: Nat.succ (n + (m + p)) = Nat.succ ((n + m) + p) := \n", - " -- 3. Now that the parentheses (apps of plus) are in the right place for both sides, push the succ on the left twice so both terms are the same.\n", - " have h_push_succ_in_left_twice: Nat.succ ((n + m) + p) = ((Nat.succ n) + m) + p := \n", - " -- Combine facts to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "4" - } -] \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_train.json b/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_train.json deleted file mode 100644 index 11b232c..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug5_sf/dsp_debug5_sf_train.json +++ /dev/null @@ -1,117 +0,0 @@ -[ - { - "nl_problem": ["Prove that for any natural number n, n + 0 = n."], - "nl_solution": [ - "Consider any natural number n.", - "Using the properties of addition, we know that adding zero to any number does not change the value of that number.", - "Therefore, we can conclude that n + 0 = n." - ], - "nl_solution_sketch": [ - "Consider any natural number n.", - "From properties of addition, adding zero does not change its values.", - "Thus, n + 0 = n." - ], - "fl_problem": ["theorem n_plus_zero_normal : ∀ n : ℕ, n + 0 = n := "], - "fl_partial_sketch": [ - "by\n", - " -- We have the fact of addition n + 0 = n, use it to show left and right are equal.\n", - " have h_nat_add_zero: ∀ n : ℕ, n + 0 = n := \n", - " -- Combine facts with to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "0" - }, - - - - { - "nl_problem": ["Prove that for any natural number n, n + (m + 1) = (n + m) + 1."], - "nl_solution": [ - "Consider any natural numbers n and m. We want to show that n + (m + 1) = (n + m) + 1.", - "Using the properties of addition, we know that adding 1 to the sum of n and m is the same as first adding m to n and then adding 1.", - "Therefore, we can conclude that n + (m + 1) = (n + m) + 1." - ], - "nl_solution_sketch": [ - "Consider any natural numbers n and m.", - "From properties of addition, adding 1 to the sum of n and m is the same as first adding m to n and then adding 1.", - "Thus, n + (m + 1) = (n + m) + 1." - ], - "fl_problem": ["theorem plus_n_Sm_proved_formal_sketch : ∀ n m : ℕ, n + (m + 1) = (n + m) + 1 := "], - "fl_partial_sketch": [ - "by\n", - " -- We have the fact of addition n + (m + 1) = (n + m) + 1, use it to show left and right are equal.\n", - " have h_nat_add_succ: ∀ n m : ℕ, n + (m + 1) = (n + m) + 1 := \n", - " -- Combine facts to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "2" - }, - - - - { - "nl_problem": ["Prove that for any natural number n and m, n + m = m + n."], - "nl_solution": [ - "Consider any natural numbers n and m. We will prove the statement by induction on n.", - "Base case: When n = 0, we need to show that 0 + m = m + 0. By the definition of addition, we have 0 + m = m and m + 0 = m. Therefore, 0 + m = m + 0.", - "Inductive step: Assume that for some natural number n, n + m = m + n. We need to show that (n + 1) + m = m + (n + 1).", - "1. Start by using the fact that (n + 1) + m = n + (m + 1) and m + (n + 1) = (m + n) + 1.", - "2. By the inductive hypothesis, we have n + m = m + n. So we can replace n + (m + 1) with (m + n) + 1.", - "3. Now, both sides have the same structure, showing that (n + 1) + m = m + (n + 1).", - "Thus, by induction, we have proved that for any natural numbers n and m, n + m = m + n." - ], - "nl_solution_sketch": [ - "Consider any natural numbers n and m. We will do induction on n.", - "Base case: 0 + m = m + 0 by properties of addition.", - "Inductive step, we have n + m = m + n. Then (n + 1) + m = (n + m) + 1 = (m + n) + 1 = m + (n + 1).", - "Thus, by induction, n + m = m + n, qed." - ], - "fl_problem": ["theorem add_comm_proved_formal_sketch : ∀ n m : ℕ, n + m = m + n := "], - "fl_partial_sketch": [ - "by\n", - " -- Consider some n and m in Nats.\n", - " intros n m\n", - " -- Perform induction on n.\n", - " induction n with\n", - " | zero =>\n", - " -- Base case: When n = 0, we need to show 0 + m = m + 0.\n", - " -- We have the fact 0 + m = m by the definition of addition.\n", - " have h_base: 0 + m = m := \n", - " -- We also have the fact m + 0 = m by the definition of addition.\n", - " have h_symm: m + 0 = m := \n", - " -- Combine facts to close goal\n", - " \n", - " | succ n ih =>\n", - " -- Inductive step: Assume n + m = m + n, we need to show succ n + m = m + succ n.\n", - " -- By the inductive hypothesis, we have n + m = m + n.\n", - " have h_inductive: n + m = m + n := \n", - " -- 1. Note we start with: Nat.succ n + m = m + Nat.succ n, so, pull the succ out from m + Nat.succ n on the right side from the addition using addition facts Nat.add_succ.\n", - " have h_pull_succ_out_from_right: m + Nat.succ n = Nat.succ (m + n) := \n", - " -- 2. then to flip m + S n to something like S (n + m) we need to use the IH.\n", - " have h_flip_n_plus_m: Nat.succ (n + m) = Nat.succ (m + n) := \n", - " -- 3. Now the n & m are on the correct sides Nat.succ n + m = Nat.succ (n + m), so let's use the def of addition to pull out the succ from the addition on the left using Nat.succ_add.\n", - " have h_pull_succ_out_from_left: Nat.succ n + m = Nat.succ (n + m) := \n", - " -- Combine facts to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "3" - } -] \ No newline at end of file diff --git a/data/debug/toy_example1_dsp/dsp_debug5_sf/readme.md b/data/debug/toy_example1_dsp/dsp_debug5_sf/readme.md deleted file mode 100644 index 284a9e7..0000000 --- a/data/debug/toy_example1_dsp/dsp_debug5_sf/readme.md +++ /dev/null @@ -1,106 +0,0 @@ -https://www.evernote.com/shard/s410/nl/75276202/2170cbbd-24a1-2d25-da32-bd8f3270d190?title=prompt%20for%20creating%20toy%20example -https://chatgpt.com/c/0ad32608-cbc9-4627-a705-786ed7421826 -I want all final responses in this format: -```json -{ - "nl_problem": ["Prove that for any natural number n, n + 0 = n."], - "nl_solution": [ - "Consider any natural number n.", - "Using the properties of addition, we know that adding zero to any number does not change the value of that number.", - "Therefore, we can conclude that n + 0 = n." - ], - "nl_solution_sketch": [ - "Consider any natural number n.", - "From properties of addition, adding zero does not change its values.", - "Thus, n + 0 = n." - ], - "fl_problem": ["theorem n_plus_zero_normal : ∀ n : ℕ, n + 0 = n :="], - "fl_partial_sketch": [ - "-- Prove that n + 0 = n via a formal proof sketch with holes to be filled\n", - "theorem n_plus_zero_proved_formal_sketch'' : ∀ n : ℕ, n + 0 = n := by\n", - " -- We have the fact of addition n + 0 = n, use it to show left and right are equal.\n", - " have h_nat_add_zero: ∀ n : ℕ, n + 0 = n := \n", - " -- Combine facts with to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ], - "path_2_file": "~/gold-ai-olympiad/lean_src_proj/lean_basics/basic_nats_using_mathlib_nats2_simp_no_rw.lean", - "fl_statement_idx": "0" - }, - - - -{ - "nl_problem": ["Prove that for any natural number n, 0 + n = n."], - "nl_solution": [ - "Consider any natural number n. We will prove the statement by induction on n.", - "Base case: When n = 0, we need to show that 0 + 0 = 0. This is true by the definition of addition.", - "Inductive step: Assume that for some natural number n, 0 + n = n. We need to show that 0 + (n + 1) = (n + 1). By the definition of addition and the inductive hypothesis, we have 0 + (n + 1) = (0 + n) + 1 = n + 1. Therefore, the statement holds for n + 1.", - "Thus, by induction, we have proved that for any natural number n, 0 + n = n." - ], - "nl_solution_sketch": [ - "Consider any natural number n, and do induction on n.", - "Base case: 0 + 0 = 0 by properties of addition.", - "Inductive step we have 0 + n = n. Then 0 + (n + 1) = (0 + n) + 1 = n + 1.", - "Where, 0 + n = n by assumption,qed." - ], - "fl_problem": ["theorem zero_plus_n_proved_formal_sketch : ∀ n : ℕ, 0 + n = n :="], - "fl_partial_sketch": [ - "-- Prove that 0 + n = n by induction via a formal proof sketch with holes to be filled\n", - "theorem zero_plus_n_proved_formal_sketch'' : ∀ n : ℕ, 0 + n = n := by\n", - " -- Consider some n in Nats.\n", - " intro n\n", - " -- Perform induction on n.\n", - " induction n with\n", - " | zero =>\n", - " -- Base case: 0 + 0 = 0\n", - " have h_base: 0 + 0 = 0 := \n", - " -- Combine facts to close goal\n", - " \n", - " | succ n ih =>\n", - " -- Inductive step: assume 0 + n = n, prove 0 + succ n = succ n\n", - " have h_inductive: 0 + Nat.succ n = Nat.succ n := \\n", - " -- Combine facts to close goal\n", - " \n" - ], - "src_header_fl_problem": ["import Mathlib.Data.Nat.Basic"], - "fl_header_sketch": [ - "import Mathlib.Data.Nat.Basic", - "import Aesop" - ] - } -``` -I want to translate the following formal proof (solution) in lean 4 to a natural language proof (solution) that a human would write (without lean code in it) and eventually make it into a concise nl_solution_sketch, like the following one: -```human_problem_solution_proof.json -"nl_problem": ["Let \\[f(x) = \\left\\{\n\\begin{array}{cl} ax+3, &\\text{ if }x>2, \\\\\nx-5 &\\text{ if } -2 \\le x \\le 2, \\\\\n2x-b &\\text{ if } x <-2.\n\\end{array}\n\\right.\\]Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper)."], -"nl_solution_sketch": ["For the piecewise function to be continuous, the cases must \"meet\" at $2$ and $-2$. For example, $ax+3$ and $x-5$ must be equal when $x=2$. This implies $a(2)+3=2-5$, which we solve to get $2a=-6 \\Rightarrow a=-3$. Similarly, $x-5$ and $2x-b$ must be equal when $x=-2$. Substituting, we get $-2-5=2(-2)-b$, which implies $b=3$. So $a+b=-3+3=\\boxed{0}$."] -``` -This is my lean 4 fl theorem (fl problem) and fl proof (fl solution): -``` --- Prove that n + (m + p) = (n + m) + p -theorem add_assoc_proved_formal_sketch : ∀ n m p : ℕ, n + (m + p) = (n + m) + p := by - -- Consider some n, m, and p in Nats. - intros n m p - -- Perform induction on n. - induction n with - | zero => - -- Base case: When n = 0, we need to show 0 + (m + p) = (0 + m) + p. - -- Using the definition of addition, 0 + (m + p) = m + p and (0 + m) + p = m + p. - simp [Nat.zero_add, Nat.zero_add] - | succ n ih => - -- Inductive step: Assume n + (m + p) = (n + m) + p, we need to show succ n + (m + p) = (succ n + m) + p. - -- proof strategy is, we move succ n out (or in) enough times then use the IH until both sides are the same. - -- 1. let's start by pulling out the scc from left side and have the entire addition inside the succ. - have h_pull_add_succ_out_from_left: Nat.succ n + (m + p) = Nat.succ (n + (m + p)) := by simp [Nat.succ_add] - -- 2. Now that we have the IH hypothesis appearing inside the left, let's apply it so we have n + (m + p) = (n + m) + p. - have h_inside_left_associates: Nat.succ (n + (m + p)) = Nat.succ ((n + m) + p) := by simp [ih] - -- 3. Now that the parenthesis (apps of plus) are on the right place for both side, push the succ on the left twice so both terms are the same. - have h_push_succ_in_left_twice: Nat.succ ((n + m) + p) = ((Nat.succ n) + m) + p := by simp [Nat.succ_add, Nat.succ_add] - -- Combining these, we get succ n + (m + p) = (succ n + m) + p. - simp [h_pull_add_succ_out_from_left, h_inside_left_associates, h_push_succ_in_left_twice] -``` -use the comments to translate the fl proof (solution) to natural language solution then use that to output a natural language concise sketch. Make the natural language proof (solution) sketch concise with the core elements of the solution proof. Do this by first outputting the natural language solution, distill it into a very concise proof sketch in natural language with only the core components. Output everything in a json code block please: \ No newline at end of file