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Proof step 5 from proof diff-unchanged (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ y, ∀ y. x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y), ∀ a0. a0 ≠ a ∧ ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ x0 \ y = a ' + x0

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%notin-01: ⊦ a ∈ y → ¬ a ∈ x \ y;%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list