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

∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ max(ins≤(a, a0 + x0)) = a, a < max(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. max(ins≤(a, x0)) = a ∨ a < max(x0) ⊦ max(ins≤(a, a0 ' + x0)) = a, a < max(a0 ' + x0)

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list