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Proof step 6 from proof max-ins-max-last (previous step)

¬ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
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)

Rule argument

Additional information

Used simplifier rules: nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%ins-y: ⊦ ¬ b < a → ins≤(a, b ' + x) = a + b + x; from specification olist