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Proof step 7 from proof in-put (previous step)

b ≠ a0, ∀ a, b, d. a ∈ ax0[b, d] ↔ a = b ∨ a ∈ ax0 ⊦ a ∈ (a0 × d0)' + ax0[b, d] ↔ a ∈ ax0 ∨ a = a0 ∨ a = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ a, b, d. a ∈ ax0[b, d] ↔ a = b ∨ a ∈ ax0 ⊦ b = a0, a = a0 ∨ a ∈ ax0[b, d] ↔ a ∈ ax0 ∨ a = a0 ∨ a = b

Rule argument

Additional information

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