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Proof step 11 from proof max-ins (previous step)

max(x0) < a0, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ x0 = [], max(a ' + a0 ' + x0) = a0, a0 < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Rule argument

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [a0 ' + x0, a] flags: rewrite, current

Additional information

Used simplifier rules: max-rec: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x));%a: ⊦ (x + y) + z = x + y + z; from specification list%app-02-ant: ⊦ ¬ x + y ≠ [] ↔ x = [] ∧ y = []; from specification list%e: ⊦ a ' ≠ []; from specification list Used terms: a0 ' + x0, a