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Proof step 5 from proof union-size-03 (previous step)

∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #((s ∪ s0) ++ a) = # s0 + #(s ++ a) - #(s0 ∩ (s ++ a)), a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #((s ∪ s0) ++ a) = # s0 + # s + 1 - #(s0 ∩ (s ++ a)), a ∈ s

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ a ∈ s → #(s ++ a) = # s + 1; (s)