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Proof step 5 from proof diff-dups (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
dups((a ' + x0) \ y), a ∈ y, ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ x0
Proof step 8:
dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ x0, a ∈ y

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y