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Proof step 12 from proof firstoc (previous step)

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

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 3 terms: (a0 × d)' + ax, d0, ax1

Additional information

Used simplifier rules: inrec: ⊦ a ∈ (b × d0)' + ax ↔ a = b ∨ a ∈ ax;%a: ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1; from specification list[assoclista] Used terms: (a0 × d)' + ax, d0, ax1