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Proof step 3 from proof union-sizeeq (previous step)

#(s ∪ s0) = # s ⊦ ∀ a. a ∈ s0 → a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
#(s ∪ s0) = # s, a ∈ s0, a0 ∈ s0 ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ ∅; (s)% ⊦ s1 = s2 ↔ (∀ a. a ∈ s1 ↔ a ∈ s2); (s)%ne-01: ⊦ a ∈ s → s ≠ ∅; (forward) from specification set-basic