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Proof step 18 from proof max-ins-max-last (previous step)

a0 < a, ordered≤(a0 ' + x), a < max(x), max(a0 ' + x) < max(x), a0 < max(x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, x) = x + a ', a < max(a0 ' + x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ 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] → [x, a0] flags: rewrite, current

Additional information

Used simplifier rules: ls-le: ⊦ a < b → (¬ b < a ↔ true); from specification oelem%ins-e: ⊦ ins≤(a, []) = a '; from specification olist%Nil: ⊦ [] + x = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c-04: ⊦ a ' = b ' ↔ a = b; from specification list Used terms: x, a0