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Proof step 13 from proof intu-empty (previous step)

a ∈ s, a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ # s = 0, s = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a ∈ s, a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s1, ∀ a. ¬ a ∈ s ∨ ¬ a ∈ s0 ⊦ # s = 0, # s1 = 0, s = ∅, s1 = ∅

Rule argument

Additional information

Used simplifier rules: size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic