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Proof step 3 from proof max-ins-max-last (previous step)

∀ a. ordered≤(x) → ins≤(a, x) = x + a ' ∨ a < max(x), ordered≤(a0 + 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 4:
ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, a0 ' + x) = (a0 ' + x) + a ', a < max(a0 ' + x)

Rule argument

Additional information

Used simplifier rules: cons-02: ⊦ ordered≤(a ' + x) → ordered≤(x); from specification olist%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list