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Proof step 8 from proof max-ins-max-last (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 10:
ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ a ' + a0 ' + x = (a0 ' + x) + a ', x = [], a0 < (max(x) < a0 ⊃ a0;max(x)), a < (max(x) < a0 ⊃ a0;max(x)), a0 < a

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: max-base: ⊦ max(a ') = a;% ⊦ x + [] = x; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ [] + x = x; (s)% ⊦ ordered≤(a '); (s)% ⊦ ¬ b < a → (a < b ↔ a ≠ b); (ss) Used terms: x, a0