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Proof step 9 from proof ins-max (previous step)

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), ordered≤(a ' + a0 ' + x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a0 < a, ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, x) = x + a ', a < max(a0 ' + x), ordered≤(a ' + a0 ' + x)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%app-01: ⊦ x + y = x + z ↔ y = z; from specification list%c: ⊦ a + x = a ' + x; from specification list