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Proof step 6 from proof lookup (previous step)

∀ a, d, ax1. (ax0 + (a × d)' + ax1)[a] = d ∨ a ∈ ax0 ⊦ a0 = a, (ax0 + (a0 × d0)' + ax1)[a0] = d0, a0 ∈ ax0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a0, d0, ax1

Additional information

Used terms: a0, d0, ax1