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Proof step 10 from proof max-append (previous step)

max(x1 + x0) < a, ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x0) ∨ max(x1 + x0) = max(x1) ⊦ x1 + x0 = [], a = max(a ' + x1), a = max(x0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 12:
max(x1 + x0) < a, ∀ x0. x1 + x0 = [] ∨ max(x1 + x0) = max(x0) ∨ max(x1 + x0) = max(x1) ⊦ x1 + x0 = [], a = max(x0), a = (max(x1) < a ⊃ a;max(x1)), x1 = []

Rule argument

spec: <toplevel> name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x1, a] flags: <all paths>

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list Used terms: x1, a