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Proof step 6 from proof append-prefix-left (previous step)

# ar1 = 0 ⊦ ar0 + ar1 = ar0

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: array name: Extension seq: ⊦ ar1 = ar2 ↔ # ar1 = # ar2 ∧ (∀ n. n < # ar1 → ar1[n] = ar2[n]) subst: [ar2, ar1] → [ar0, ar0 + ar1] flags: <all paths>

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[n];% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n + 0 = n; (s) Used terms: ar0, ar0 + ar1