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Proof step 5 from proof max-ins-last (previous step)

max(a0 ' + x0) < a, ∀ a. max(x0) < a → ins≤(a, x0).last = a ⊦ ins≤(a, a0 ' + x0).last = a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 7:
(max(x0) < a0 ⊃ a0;max(x0)) < a, ∀ a. max(x0) < a → ins≤(a, x0).last = a ⊦ ins≤(a, a0 ' + x0).last = a, x0 = []

Rule argument

spec: <toplevel> name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x0, a0] flags: <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ b < a → ins≤(a, b ') = b ' + a '; (s)% ⊦ (x + a ').last = a; (s) Used terms: x0, a0