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Proof step 8 from proof diff-unchanged (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 9:
x0 \ y = x0, ∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y) ⊦ x0 \ y = x0 ↔ (∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y), a ∈ y
Proof step 13:
∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y) ⊦ x0 \ y = x0, x0 \ y = x0 ↔ (∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y), a ∈ y

Rule argument

fma: x0 \ y = x0

Additional information

Used formulas: x0 \ y = x0