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

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

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, a < (max(x1) < a ⊃ a;max(x1)), max(x1 + x0) < (max(x1) < a ⊃ a;max(x1)) ⊦ x1 + x0 = [], x1 = [], ∃ x0. max(x1 + x0) < max(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 terms: x1, a