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Proof step 13 from proof max-ins-max (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 15:
a0 < a, ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ (max(ins≤(a, x0)) < a0 ⊃ a0;max(ins≤(a, x0))) = a, ins≤(a, x0) = [], a < max(a0 ' + x0)

Rule argument

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

Additional information

Used simplifier rules: ⊦ ins≤(a, x) ≠ []; (s) Used terms: ins≤(a, x0), a0