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Proof step 4 from proof cons-01 (previous step)

disj(x, y0), disj(a ' + x, y0), a ∈ y0, ∀ x0, a1. ¬ disj(x0, y0) ∨ ¬ a1 ∈ y0 ∨ ¬ disj(a1 ' + x0, y0) ⊦ a0 = a, a0 ∈ x

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: x, a

Additional information

Used terms: x, a