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Proof step 3 from proof in-17 (previous step)

⊦ (∃ y, z. b ' = y + (a + z)) ↔ a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∃ y, z. b ' = y + a ' + z) ↔ a = b

Rule argument

Additional information

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