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Proof step 7 from proof last-ex (previous step)

∃ y, z. y0 = y + (a + z) ⊦ a = a0, ∃ y, z. y0 + a0 ' = y + a ' + z ∧ z ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ a = a0, ∃ y0, z0. y + (a ' + z) + a0 ' = y0 + a ' + z0 ∧ z0 ≠ []

Rule argument

Additional information

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