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Proof step 2 from proof union-size-02 (previous step)

a ∈ s0 ⊦ #(s ∪ s0) = #(s ∪ (s0 -- a)) + 1, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ s0, a0 ∈ s0 ⊦ #(s ∪ s0) = #((s ∪ s0) -- a) + 1, 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;% ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic