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Proof step 5 from proof max-ge-every (previous step)

max(a0 ' + x0) < a, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ a = a0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 7:
(max(x0) < a0 ⊃ a0;max(x0)) < a, a ∈ x0, ∀ a1. ¬ max(x0) < a1 ∨ ¬ a1 ∈ x0 ⊦ a = a0, x0 = []

Rule argument

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

Additional information

Used simplifier rules: ⊦ ¬ a ∈ []; (s) Used terms: x0, a0