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Proof step 6 from proof ex (previous step)

∃ y, z. ins≤(a, y + z) = y + a ' + z ∧ x0 = y + z, ∀ 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 7:
ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

Proof step 6 from proof ex (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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