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Proof step 1 from proof inapp-01 (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z;%inapp-03: ⊦ ¬ a ∈ y ∧ ¬ b ∈ x → (x + a ' + z = y + b ' + y0 ↔ a = b ∧ x = y ∧ z = y0);