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

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a

Additional information

Used simplifier rules: Difference: ⊦ a ∈ s1 \ s2 ↔ a ∈ s1 ∧ ¬ a ∈ s2; Used terms: a