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Proof step 4 from proof diff-unchanged (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a ∈ y, ∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y) ⊦ (a ' + x0) \ y = a ' + x0 ↔ (∀ a0. a0 ≠ a ∧ ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y)
Proof step 7:
∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y) ⊦ a ∈ y, (a ' + x0) \ y = a ' + x0 ↔ (∀ a0. a0 ≠ a ∧ ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y)

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y