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Proof step 1 from proof inapp (conclusion)

⊦ ¬ a ∈ x ∧ ¬ a ∈ y → (x + a ' + y = z + a ' + y0 ↔ y = y0 ∧ x = z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' + y = z + a ' + y0 ⊦ a ∈ y, a ∈ x, y = y0 ∧ x = z

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;