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Proof step 15 from proof subset-is-union (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
s0 ∩ s = ∅ ++ a, a ∈ s ⊦ # s = 0, s = ∅, a ∈ s0

Rule argument

Additional information

Used simplifier rules: int-c: ⊦ s ∩ s0 = s0 ∩ s;%del-e-02: ⊦ s -- a = ∅ ↔ s = ∅ ∨ s = ∅ ++ a;%size-one-01: ⊦ #(∅ ++ a) = 1; from specification set-basic%ne: ⊦ ∅ ≠ s ++ a; from specification set-basic%size-zero: ⊦ # s ≠ 0 → s ≠ ∅; (forward) from specification set-basic%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%one-size: ⊦ s = ∅ ++ a → # s = 1; (forward) from specification set-basic