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Proof step 15 from proof diff-in (previous step)

a ∈ x0 \ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0, a ∈ x0, a0 ∈ y

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 16:
a ∈ x0 \ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0, a = a0, a ∈ x0, a0 ∈ y

Rule argument

fma: a = a0

Additional information

Used formulas: a = a0