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Proof step 3 from proof diff-in (previous step)

∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ (a0 ' + x0) \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a0 ∈ y, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ (a0 ' + x0) \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y
Proof step 10:
∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a0 ∈ y, a ∈ (a0 ' + x0) \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y

Rule argument

fma: a0 ∈ y

Additional information

Used formulas: a0 ∈ y