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Proof step 11 from proof diff-in (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 12:
a ∈ x0, ∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a = a0 ∨ a ∈ x0 \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y, a0 ∈ y
Proof step 14:
∀ a, y. a ∈ x0 \ y ↔ a ∈ x0 ∧ ¬ a ∈ y ⊦ a ∈ x0, a = a0 ∨ a ∈ x0 \ y ↔ (a = a0 ∨ a ∈ x0) ∧ ¬ a ∈ y, a0 ∈ y

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0