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Proof step 4 from proof max-ins-max-last (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x)
Proof step 6:
¬ a0 < a, ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, a0 ' + x) = (a0 ' + x) + a ', a < max(a0 ' + x)
Proof step 15:
a0 < a, ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ a0 + ins≤(a, x) = (a0 ' + x) + a ', a < max(a0 ' + x)

Rule argument

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

Additional information

Used terms: x, a0, a