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Proof step 2 from proof copy-exact (previous step)

# ar1 = # ar0 ⊦ copy(ar0, 0, ar1, 0, # ar0) = ar0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n])

Rule argument

spec: array inst: name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ar0, copy(ar0, 0, ar1, 0, # ar0)] flags: rewrite

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%at-copy: ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]);%zero-01: ⊦ ¬ m < 0; from specification nat-basic1%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%sub-base: ⊦ m - 0 = m; from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat Used terms: ar0, copy(ar0, 0, ar1, 0, # ar0)