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Proof step 8 from proof put-in (previous step)

a ≠ a1, ∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 ∈ (a1 × d1)' + ax0[a, d] ↔ a0 × d0 = a × d, a0 × d0 = a1 × d1, a0 × d0 ∈ ax0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
∀ a1, d1, a2, d2. (a1 × d1 ∈ ax0[a2, d2] ↔ a1 = a2 ∧ d1 = d2) ∨ a1 × d1 ∈ ax0 ⊦ a0 × d0 = a1 × d1, a = a1, a0 × d0 ∈ ax0[a, d] ↔ a0 × d0 = a × d, a0 × d0 ∈ ax0

Rule argument

Additional information

Used simplifier rules: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ p ≠ p0 → (p ∈ p0 ' + ax ↔ p ∈ ax);