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Proof step 1 from proof array-le-same-prepart (conclusion) (back to tree)

⊦ ar1 ≤ ar2 ∧ ar2 ≤ ar3 ∧ subarray(ar1, 0, m) = subarray(ar3, 0, m) ∧ m < # ar1 ∧ m < # ar2 ∧ m < # ar3 → subarray(ar1, 0, m) = subarray(ar2, 0, m)

System applied rule normalize

And obtained the following new goals

Rule argument

Additional information

Proof step 2 from proof array-le-same-prepart (previous step) (back to tree)

subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar1, 0, m) ≠ subarray(ar2, 0, m), m < # ar1, m < # ar2, m < # ar3, ar1 ≤ ar2, ar2 ≤ ar3 ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:array-split-into-subarrays, seq:m < # ar ⊦ ar = subarray(ar, 0, m) + subarray(ar, m, # ar - m), subst:[m, ar] → [m, ar3]

Additional information

Used terms: m, ar3

Proof step 3 from proof array-le-same-prepart (previous step) (back to tree)

m < # ar ⊦ ar = subarray(ar, 0, m) + subarray(ar, m, # ar - m)

Lemma array-split-into-subarrays

Proof step 4 from proof array-le-same-prepart (previous step) (back to tree)

ar3 = subarray(ar3, 0, m) + subarray(ar3, m, # ar3 - m), subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar1, 0, m) ≠ subarray(ar2, 0, m), m < # ar3, m < # ar2, m < # ar1, ar2 ≤ ar3, ar1 ≤ ar2 ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:array-split-into-subarrays, seq:m < # ar ⊦ ar = subarray(ar, 0, m) + subarray(ar, m, # ar - m), subst:[m, ar] → [m, ar1]

Additional information

Used terms: m, ar1

Proof step 5 from proof array-le-same-prepart (previous step) (back to tree)

m < # ar ⊦ ar = subarray(ar, 0, m) + subarray(ar, m, # ar - m)

Lemma array-split-into-subarrays

Proof step 6 from proof array-le-same-prepart (previous step) (back to tree)

ar1 = subarray(ar1, 0, m) + subarray(ar1, m, # ar1 - m), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, # ar3 - m), subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar1, 0, m) ≠ subarray(ar2, 0, m), m < # ar1, m < # ar2, m < # ar3, ar1 ≤ ar2, ar2 ≤ ar3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: array-le-trans: ⊦ ar0 ≤ ar1 ∧ ar1 ≤ ar2 → ar0 ≤ ar2; ⊦ # x = 0 ↔ x = []; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic

Proof step 7 from proof array-le-same-prepart (previous step) (back to tree)

ar1 = subarray(ar3, 0, m) + subarray(ar1, m, # ar1 - m), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, # ar3 - m), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar1, m < # ar2, m < # ar3, ar1 ≤ ar3, ar1 ≤ ar2, ar2 ≤ ar3 ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [# ar1, m, n0]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar1, m, n0

Proof step 8 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, # ar3 - m), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar3, m < # ar2, m < # ar1, ar2 ≤ ar3, ar1 ≤ ar2, ar1 ≤ ar3, m ≤ # ar1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < m + n ↔ n ≠ 0; (s) ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 9 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, # ar3 - m), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], n0 ≠ 0, subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar2, m < # ar3, ar1 ≤ ar3, ar1 ≤ ar2, ar2 ≤ ar3 ⊦

Heuristic elimination applied rule apply elim lemma

And obtained the following new goals

Rule argument

spec: nat, name:elim-sub, seq:n ≤ m ⊦ n0 = m - n ↔ m = n0 + n, subst:[m, n, n0] → [# ar3, m, n1]

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: # ar3, m, n

Proof step 10 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, # ar3 = n1 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], n0 ≠ 0, subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar3, m < # ar2, ar2 ≤ ar3, ar1 ≤ ar2, ar1 ≤ ar3, m ≤ # ar3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ m < m + n ↔ n ≠ 0; (s) ⊦ m < n → n ≠ 0; (s) lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic ⊦ m < n → m ≤ n; (s)

Proof step 11 from proof array-le-same-prepart (previous step) (back to tree)

# ar3 = n1 + m, # ar1 = n0 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], n0 ≠ 0, n1 ≠ 0, subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar2, ar1 ≤ ar3, ar1 ≤ ar2, ar2 ≤ ar3 ⊦

User applied rule apply lemma

And obtained the following new goals

Rule argument

spec: <toplevel>, name:array-le-same-prefix, seq: ⊦ x + y1 ≤ ar2 ∧ ar2 ≤ x + y3 → (∃ y2. ar2 = x + y2), subst:[x, y1, ar2, y3] → [subarray(ar3, 0, m), subarray(ar1, m, n0), ar2, subarray(ar3, m, n1)]

Additional information

Used terms: subarray(ar3, 0, m), subarray(ar1, m, n0), ar2, subarray(ar3, m, n1)

Proof step 12 from proof array-le-same-prepart (previous step) (back to tree)

⊦ x + y1 ≤ ar2 ∧ ar2 ≤ x + y3 → (∃ y2. ar2 = x + y2)

Lemma array-le-same-prefix

Proof step 13 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, # ar3 = n1 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), ar1 ≠ [], ar2 ≠ [], ar3 ≠ [], n0 ≠ 0, n1 ≠ 0, subarray(ar3, 0, m) ≠ subarray(ar2, 0, m), m < # ar2, ar2 ≤ ar3, ar1 ≤ ar2, ar1 ≤ ar3, subarray(ar3, 0, m) + subarray(ar1, m, n0) ≤ ar2 ∧ ar2 ≤ subarray(ar3, 0, m) + subarray(ar3, m, n1) → (∃ y2. ar2 = subarray(ar3, 0, m) + y2) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ # x = 0 ↔ x = []; (s) ⊦ m < m + n ↔ n ≠ 0; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ # subarray(ar, m, n) = n; (s) ⊦ #(x + y) = # x + # y; (s)

Proof step 14 from proof array-le-same-prepart (previous step) (back to tree)

# ar3 = n1 + m, # ar1 = n0 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar3, 0, m) + ar ≠ [], ar ≠ [], ar1 ≠ [], ar3 ≠ [], n0 ≠ 0, n1 ≠ 0, subarray(ar3, 0, m) ≠ subarray(subarray(ar3, 0, m) + ar, 0, m), ar1 ≤ ar3, ar1 ≤ subarray(ar3, 0, m) + ar, subarray(ar3, 0, m) + ar ≤ ar3 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: array[natarray], name:Extension, seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]), subst:[ar1, ar2] → [subarray(ar3, 0, m), subarray(subarray(ar3, 0, m) + ar, 0, m)], <all paths>

Additional information

Used terms: subarray(ar3, 0, m), subarray(subarray(ar3, 0, m) + ar, 0, m)

Proof step 15 from proof array-le-same-prepart (previous step) (back to tree)

⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])

A spec-lemma

Proof step 16 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, # ar3 = n1 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar3, 0, m) + ar ≠ [], ar ≠ [], ar1 ≠ [], ar3 ≠ [], n0 ≠ 0, n1 ≠ 0, subarray(ar3, 0, m) + ar ≤ ar3, ar1 ≤ subarray(ar3, 0, m) + ar, ar1 ≤ ar3, # subarray(ar3, 0, m) ≠ # subarray(subarray(ar3, 0, m) + ar, 0, m) ∨ (∃ n. ¬(n < # subarray(ar3, 0, m) → subarray(ar3, 0, m)[n] = subarray(subarray(ar3, 0, m) + ar, 0, m)[n])) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s) f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic ⊦ n + 0 = n; (s) ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) ⊦ # subarray(ar, m, n) = n; (s)

Proof step 17 from proof array-le-same-prepart (previous step) (back to tree)

# ar1 = n0 + m, # ar3 = n1 + m, ar1 = subarray(ar3, 0, m) + subarray(ar1, m, n0), ar3 = subarray(ar3, 0, m) + subarray(ar3, m, n1), subarray(ar1, 0, m) = subarray(ar3, 0, m), subarray(ar3, 0, m) + ar ≠ [], ar3[n] ≠ (subarray(ar3, 0, m) + ar)[n], ar ≠ [], ar1 ≠ [], ar3 ≠ [], n0 ≠ 0, n1 ≠ 0, n < m, subarray(ar3, 0, m) + ar ≤ ar3, ar1 ≤ subarray(ar3, 0, m) + ar, ar1 ≤ ar3 ⊦

User applied rule apply rewrite lemma

And obtained the following new goals

Rule argument

spec: list-del[natarray-oarray-sort], name:get-append-less, seq: ⊦ n < # x → (x + y)[n] = x[n], subst:[x, y, n] → [subarray(ar3, 0, m), ar, n], <all paths>

Additional information

Used simplifier rules: ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s) ⊦ n + 0 = n; (s) a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic c: ⊦ m + n = n + m; from specification nat-basic ⊦ ¬ n < # x → (x + y)[n] = y[n - # x]; (s) a: ⊦ (x + y) + z = x + y + z; from specification list[natarray-oarray-sort] ⊦ # subarray(ar, m, n) = n; (s) ⊦ m < n ↔ ¬ n ≤ m; (ss) ⊦ m < n → (n ≤ m ↔ false); (s) Used terms: subarray(ar3, 0, m), ar, n

Proof step 18 from proof array-le-same-prepart (previous step) (back to tree)

⊦ n < # x → (x + y)[n] = x[n]

A spec-lemma

Proof Tree for array-le-same-prepart (View conclusion)

The following image shows the proof tree of sequent calculus. The proved goal is at the bottom. You can click on the nodes to view additional rules. See here for details on colors in a proof tree.

the proof tree