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Proof step 12 from proof rem-inter-empty3 (previous step)

a ∈ s2, a ∈ s, ¬ a ∈ s2 \ s0 ∨ ¬ a ∈ s, ∀ a0. ¬ a0 ∈ s0 \ s1 ∨ ¬ 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 13:
a ∈ s0, a ∈ s, a ∈ s2, ∀ a0. ¬ a0 ∈ s0 \ s1 ∨ ¬ a0 ∈ s ⊦ s0 = ∅, # s = 0, # s2 = 0, #((s2 \ s1) ∩ s) = 0, s = ∅, s2 = ∅, # s0 = 0, a ∈ s1

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