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Proof step 3 from proof bufs-eq (previous step)

(σ to n).buf = (σ to m).buf ⊦ n = m, # σ < ⌜n⌝ , # σ < ⌜m⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n])

Rule argument

spec: array, inst: buffer name: Extension seq: ⊦ buf = buf0 ↔ # buf = # buf0 ∧ (∀ n. n < # buf → buf[n] = buf0[n]) subst: [buf0, buf] → [(σ to m).buf, (σ to n).buf] flags: <all paths>

Additional information

Used simplifier rules: ⊦ finp(σ0 to m); (s)%finp(σ) ⊦ # σ.buf = (# σ).n; (s)%¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝ ; (s)% ⊦ ⌜m⌝ .n = m; (s) Used terms: (σ to m).buf, (σ to n).buf