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Proof step 7 from proof eq (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
ins≤(a, a0 ' + x0) = ins≤(a, y), ∀ a, y. ins≤(a, x0) = ins≤(a, y) → x0 = y ⊦ a0 ' + x0 = y

Rule argument

Additional information

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