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Proof step 1 from proof ex (conclusion)

⊦ a ∈ ax ↔ (∃ d, ay, az. ax = ay + (a × d)' + az)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ ax ⊦ ∃ d, ay, az. ax = ay + (a × d)' + az

Rule argument

Additional information

Used simplifier rules: inapp: ⊦ a ∈ ax + ay ↔ ¬ ¬(a ∈ ax ∨ a ∈ ay);%inone: ⊦ a ∈ (b × d0)' ↔ a = b;%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista]