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Proof step 4 from proof copy-exact-append (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ 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 + ar1, 0, ar, 0, # ar)] flags: rewrite

Additional information

[Concretesimprules((Csimpnamedseq(append-at1,,, ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n]) Csimpnamedseq(append-a,,, ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1) Csimpnamedseq(array-copy,,size-copy, ⊦ # copy(ar, m, ar0, m0, n) = # ar0) Csimpnamedseq(array-copy,,at-copy, ⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0] ;ar0[m0 + n - n0] )) Csimpnamedseq(nat-basic1,,zero-01, ⊦ ¬ m < 0) Csimpnamedseq(nat-basic2,,add-base, ⊦ n + 0 = n) Csimpnamedseq(nat-basic2,,a, ⊦ (m + n0) + n1 = m + n0 + n1) Csimpnamedseq(nat-basic2,,c, ⊦ m + n = n + m) Csimpnamedseq(nat,,ls-le-simp, ⊦ m < n → (n ≤ m ↔ false)) Csimpnamedseq(nat,,sub-base, ⊦ m - 0 = m) Csimpnamedforward(nat-basic1,,lf-01, ⊦ m < n ∧ n = n0 → m < n0) Csimpnamedforward(nat-basic1,,f, ⊦ m < n → n ≠ 0))), Extraterms((ar0copy(ar0 + ar1, 0, ar, 0, # ar)))]