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Proof step 4 from proof rem-inter-empty3 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
(s0 \ s1) ∩ s = ∅, (s2 \ s0) ∩ s = ∅, a ∈ s2, a ∈ 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;%Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic%In-empty: ⊦ ¬ a ∈ ∅; from specification set-basic