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Proof step 12 from proof firstoc (previous step)

⊦ a = a0, a0 ∈ ax, ∃ ax2, d1, ax3. ¬ a0 ∈ ax2 ∧ (a × d)' + ax + (a0 × d0)' + ax1 = ax2 + (a0 × d1)' + ax3

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: (a × d)' + ax, d0, ax1

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; Used terms: (a × d)' + ax, d0, ax1