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Proof step 6 from proof diff-empty-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ y. x0 \ y = [] ↔ (∀ a. a ∈ x0 → a ∈ y), ∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y ⊦ a ∈ y

Rule argument

Additional information

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