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Proof step 10 from proof rem-inter-empty3 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a ∈ s2, a ∈ s, ∀ a0. ¬ a0 ∈ s2 \ s0 ∨ ¬ a0 ∈ s, ∀ a0. ¬ a0 ∈ s0 \ s1 ∨ ¬ a0 ∈ s ⊦ s2 = ∅, s = ∅, #((s2 \ s1) ∩ s) = 0, # s2 = 0, # s = 0, a ∈ s1

Rule argument

Additional information

Used simplifier rules: int-rax: ⊦ a ∈ s1 ∩ s2 ↔ ¬ ¬(a ∈ s1 ∧ a ∈ s2);%int-a: ⊦ (s ∩ s0) ∩ s1 = s ∩ s0 ∩ s1;%int-c: ⊦ s ∩ s0 = s0 ∩ s;%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic