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Proof step 13 from proof rem-inter-empty3 (previous step)

a ∈ s1, a ∈ s2, ∀ a0. ¬ a0 ∈ s \ s0 ∨ ¬ a0 ∈ s1, ¬ a ∈ s2 \ s ∨ ¬ a ∈ s1 ⊦ # 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 14:
a ∈ s, a ∈ s1, a ∈ s2, ∀ a0. ¬ a0 ∈ s \ s0 ∨ ¬ a0 ∈ s1 ⊦ # s1 = 0, # s2 = 0, #((s2 \ s0) ∩ s1) = 0, # s = 0, s = ∅, s1 = ∅, s2 = ∅, a ∈ s0

Rule argument

Additional information

Used simplifier rules: 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