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Proof step 9 from proof in-def (previous step)

a ∈ x0, a ∈ x0 → (∃ y, z. x0 = y + a ' + z) ⊦ a = a0, ∃ y, z. a0 ' + x0 = y + a ' + z

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
⊦ a = a0, ∃ y0, z0. a0 ' + y + a ' + z = y0 + a ' + z0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%in-13: ⊦ x = y + a ' + z → a ∈ x;