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Proof step 13 from proof ins-max (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 14:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))

Rule argument

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

Additional information

Used simplifier rules: ⊦ a < b → (b < a ↔ false); (ss)% ⊦ ins≤(a, []) = a '; (s)% ⊦ [] + x = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ' = b ' ↔ a = b; (s) Used terms: x, a0