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Proof step 8 from proof firstoc (previous step)

∀ a. a ∈ ax0 → (∃ ax2, d, ax1. ax0 = ax2 + (a × d)' + ax1 ∧ ¬ a ∈ ax2) ⊦ ∃ ax, d0, ax1. ¬ a0 ∈ ax ∧ (a0 × d)' + ax0 = ax + (a0 × d0)' + ax1

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 1 terms: [], d, ax0

Additional information

Used simplifier rules: empty: ⊦ ¬ a ∈ [];% ⊦ [] + ax = ax;% ⊦ (ax + ax0) + ax1 = ax + ax0 + ax1;% ⊦ ax + ax0 = ax + ax1 ↔ ax0 = ax1; Used terms: [], d, ax0