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Proof step 6 from proof in-rec (previous step)

⊦ (∃ y, z. b ' + x = y + (a + z)) ↔ a = b ∨ (∃ y, z. x = y + (a + z))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
⊦ (∃ y, z. b ' + x = y + a ' + z) ↔ a = b ∨ (∃ y, z. x = y + a ' + z)

Rule argument

Additional information

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