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Proof step 14 from proof in-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ x = y + b ' + z, ∃ y, z. b ' + x = y + b ' + z

Rule argument

Additional information

Used simplifier rules: app-08: ⊦ a ' + x = a ' + y ↔ x = y;%a: ⊦ (x + y) + z = x + y + z;