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Proof step 3 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)

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a ∈ y, ∀ y. x0 \ y = [] ↔ (∀ a. a ∈ x0 → a ∈ y) ⊦ (a ' + x0) \ y = [] ↔ (∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y)
Proof step 6:
∀ y. x0 \ y = [] ↔ (∀ a. a ∈ x0 → a ∈ y) ⊦ a ∈ y, (a ' + x0) \ y = [] ↔ (∀ a0. a0 = a ∨ a0 ∈ x0 → a0 ∈ y)

Rule argument

fma: a ∈ y

Additional information

Used formulas: a ∈ y