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Proof step 28 from proof max-ins-max (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 29:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 30:
a0 < a, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ max(ins≤(a, x0)) = a, x0 = [], a < (max(x0) < a0 ⊃ a0;max(x0)), max(ins≤(a, x0)) < a0

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;% ⊦ ins≤(a, []) = a '; (s) Used terms: x0, a0