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Proof step 15 from proof ins-max (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x))
Proof step 17:
x ≠ [], ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ a ' + a0 ' + x = (a0 ' + x) + a ', a < (max(x) < a0 ⊃ a0;max(x)), a0 < a, a0 < (max(x) < a0 ⊃ a0;max(x))

Rule argument

spec: inst: name: max-rec seq: ⊦ x ≠ [] → max(a ' + x) = (max(x) < a ⊃ a;max(x)) subst: [x, a] → [x, a0] flags: rewrite, current

Additional information

Used simplifier rules: max-base: ⊦ max(a ') = a;%app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%le-o: ⊦ ordered≤(a '); from specification olist%seq: ⊦ ¬ b < a → (¬ a < b ↔ a = b); from specification oelem Used terms: x, a0