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Proof step 1 from proof eq-03 (conclusion)

⊦ x = y + a ' ∧ x = z + b ' → a = b ∧ y = z

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b;%a: ⊦ (x + y) + z = x + y + z;