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Proof step 1 from proof inapp-03 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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