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Proof step 11 from proof subseteq-inter-empty-empty (previous step)

a ∈ s2, a ∈ s0, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2, a ∈ s0 → a ∈ s1, ∀ a. a ∈ s0 → a ∈ s1 ⊦ s2 = ∅, s0 = ∅, #(s0 ∩ s2) = 0, # s2 = 0, # s0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ s1, a ∈ s0, a ∈ s2, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2, ∀ a. a ∈ s0 → a ∈ s1 ⊦ s1 = ∅, # s0 = 0, # s2 = 0, #(s0 ∩ s2) = 0, s0 = ∅, s2 = ∅, # s1 = 0

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