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Proof step 4 from proof diff-dups (previous step)

dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ a ∈ x0, dups(x0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a ∈ y, dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ a ∈ x0, dups(x0)
Proof step 7:
dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ a ∈ y, a ∈ x0, dups(x0)

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y