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Proof step 6 from proof max-ins (previous step)

⊦ max(a ' + a0 ') = a0, a0 < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ 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 ', a] flags: rewrite, current

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%e: ⊦ a ' ≠ []; from specification list Used terms: a0 ', a