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Proof step 1 from proof union-size-02 (conclusion)

⊦ ¬ a ∈ s ∧ a ∈ s0 → #(s ∪ s0) = #(s ∪ (s0 -- a)) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ s0 ⊦ s0 = ∅, #(s ∪ s0) = #((s ∪ s0) -- a) + 1, # s0 = 0, a ∈ s

Rule argument

Additional information

Used simplifier rules: undel-03: ⊦ ¬ a ∈ s → s ∪ (s0 -- a) = (s ∪ s0) -- a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%size-zero-01: ⊦ s ≠ ∅ → # s ≠ 0; (forward) from specification set-basic%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic