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Proof step 11 from proof ex (previous step)

¬ a0 < a, ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ a + a0 + (y + z) = (a0 ' + y) + a ' + z, ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
ins≤(a, y + z) = y + a ' + z, ∀ a. ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ y + z = y0 + z0 ⊦ a ' + a0 ' + y = a0 ' + y + a ', a0 < a, ∃ y0, z0. ins≤(a, y0 + z0) = y0 + a ' + z0 ∧ a0 ' + y + z = y0 + z0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%app-03: ⊦ x + y = z + y ↔ x = z; from specification list%c: ⊦ a + x = a ' + x; from specification list