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Proof step 6 from proof insfirst (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a, d, ax0, d0

Additional information

Used terms: a, d, ax0, d0