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Proof step 12 from proof subset-is-union (previous step)

(s -- a) ∩ s0 = ∅, a ∈ s ⊦ s = ∅, # s = 0, a ∈ s0, s ∪ s0 = s0 ∪ s ∧ s ∩ s0 = ∅

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
(s -- a) ∩ s0 = ∅, a ∈ s ⊦ #(s ∩ s0) = 0, s ∩ s0 = ∅, # s = 0, s = ∅, a ∈ s0

Rule argument

Additional information

Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s;%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic