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Proof step 4 from proof in-17 (previous step)

⊦ (∃ y, z. b ' = y + a ' + z) ↔ a = b

User applied rule case distinction

And obtained the following new goals

Proof step 5:
a = b ⊦ ∃ y, z. b ' = y + a ' + z

Rule argument

pos: right 1

Additional information

Used simplifier rules: eq-04: ⊦ x + a ' + y = b ' ↔ x = [] ∧ a = b ∧ y = [];%a: ⊦ (x + y) + z = x + y + z;