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

∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(ins≤(a, a0 ' + x0)) = max(x0), x0 = [], max(x0) < a0, max(x0) < a

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 16:
⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x)
Proof step 17:
∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(ins≤(a, a0 ' + x0)) = max(x0), x0 = [], max(x0) < a, max(x0) < a0, a0 < a
Proof step 20:
a0 < a, ∀ a1. max(ins≤(a1, x0)) = max(x0) ∨ max(x0) < a1 ⊦ max(a0 + ins≤(a, x0)) = max(x0), x0 = [], max(x0) < a, max(x0) < a0

Rule argument

spec: olist name: ins-n seq: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x) subst: [x, b, a] → [x0, a0, a] flags: <all paths>

Additional information

Used terms: x0, a0, a