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Proof step 1 from proof inapp-06 (conclusion)

⊦ ¬ b ∈ y ∧ ¬ a ∈ z → (x + a ' + y = x0 + b ' + z ↔ x = x0 ∧ a = b ∧ y = z)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
x + a ' + y = x0 + b ' + z ⊦ a ∈ z, b ∈ y, x = x0 ∧ a = b ∧ y = z

Rule argument

Additional information

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