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Proof step 11 from proof rem-inter-empty3 (previous step)

a ∈ s1, a ∈ s2, ∀ a0. ¬ a0 ∈ s2 \ s ∨ ¬ a0 ∈ s1, ∀ a. a ∈ (s \ s0) ∩ s1 ↔ a ∈ ∅ ⊦ # s1 = 0, # s2 = 0, #((s2 \ s0) ∩ s1) = 0, s1 = ∅, s2 = ∅, a ∈ s0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ s2, a ∈ s1, ∀ a0. ¬ a0 ∈ s2 \ s ∨ ¬ a0 ∈ s1, ∀ a0. ¬ a0 ∈ s \ s0 ∨ ¬ a0 ∈ s1 ⊦ #((s2 \ s0) ∩ s1) = 0, # s2 = 0, # s1 = 0, s1 = ∅, s2 = ∅, a ∈ s0

Rule argument

Additional information

Used simplifier rules: Intersect: ⊦ a ∈ s1 ∩ s2 ↔ a ∈ s1 ∧ a ∈ s2;%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;% ⊦ ¬ a ∈ ∅; (s)