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Proof step 3 from proof rev-disjoint-01 (previous step)

∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ disj(rev(x), a ' + y0) ↔ disj(x, a ' + y0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
disj(x, a ' + y0), ∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ disj(rev(x), a ' + y0) ↔ disj(x, a ' + y0)
Proof step 6:
∀ x. disj(rev(x), y0) ↔ disj(x, y0) ⊦ disj(x, a ' + y0), disj(rev(x), a ' + y0) ↔ disj(x, a ' + y0)

Rule argument

fma: disj(x, a ' + y0)

Additional information

Used formulas: disj(x, a ' + y0)