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Proof step 1 from proof c-06 (conclusion)

⊦ a ' + x = b ' + y ↔ a = b ∧ x = y

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: One: ⊦ a ' = a + [];%Cons: ⊦ (a + x) + y = a + (x + y);%Nil: ⊦ [] + x = x;% ⊦ a + x = a0 + x0 ↔ a = a0 ∧ x = x0; (ws)