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Proof step 1 from proof firstoc-rew (conclusion)

⊦ a ∈ x ↔ (∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x ⊦ ∃ y, z. x = y + a ' + z ∧ ¬ a ∈ y

Rule argument

Additional information

Used simplifier rules: neq: ⊦ ¬ a ∈ x → x ≠ y + a ' + z;%a: ⊦ (x + y) + z = x + y + z;