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Proof step 8 from proof sw (previous step)

∀ a, b. ins≤(a, ins≤(b, x0)) = ins≤(b, ins≤(a, x0)) ⊦ ins≤(a, ins≤(b, a0 + x0)) = ins≤(b, ins≤(a, a0 + x0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a, b. ins≤(a, ins≤(b, x0)) = ins≤(b, ins≤(a, x0)) ⊦ ins≤(a, ins≤(b, a0 ' + x0)) = ins≤(b, ins≤(a, a0 ' + x0))

Rule argument

Additional information

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