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Proof step 2 from proof inapp-03 (previous step)

⊦ x + a ' + z = y + b ' + y0 ↔ a = b ∧ x = y ∧ z = y0, b ∈ x, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
x + a ' + z = y + b ' + y0 ⊦ a ∈ y, b ∈ x, a = b ∧ x = y ∧ z = y0

Rule argument

Additional information

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