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Proof step 7 from proof postfix-rec-right (previous step)

∃ z. z + x = a ' + y ⊦ x = a ' + y, ∃ z. z + x = y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
z + x = a ' + y, a ∈ z ∨ a ∈ x ⊦ z = [], ∃ z. z + x = y

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ x = y + x ↔ y = []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list