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Proof step 4 from proof diff-empty-all (previous step)

a ∈ y, ∀ y. x0 \ y = [] ↔ (∀ a. a ∈ x0 → a ∈ y) ⊦ (a ' + x0) \ y = [] ↔ (∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ y, ∀ y. x0 \ y = [] ↔ (∀ a. a ∈ x0 → a ∈ y) ⊦ x0 \ y = [] ↔ (∀ a0. a0 ∈ x0 → a0 ∈ y)

Rule argument

Additional information

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