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Proof step 34 from proof max-ins-max (previous step)

a0 ' + x0 ≠ [], ∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) = a, a0 < a, a < max(a0 ' + x0), ordered≤(a0 ' + x0), a0 < max(a0 ' + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ (max(a0 ' + x0) < a ⊃ a;max(a0 ' + x0)) = a, a0 < a, a < max(a0 ' + x0), ordered≤(a0 ' + x0), a0 < max(a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list