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Proof step 4 from proof ex (previous step)

∀ a. ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z ⊦ ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ a0 + x0 = y + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a. ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z ⊦ ∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ a0 ' + x0 = y + z

Rule argument

Additional information

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