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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)% ⊦ a + x = a ' + x; (s)