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Proof step 2 from proof union-size-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #((s ∪ s0) ++ a) = # s0 + #(s ++ a) - #(s0 ∩ (s ++ a))

Rule argument

Additional information

Used simplifier rules: unadd: ⊦ (s ++ a) ∪ s0 = (s ∪ s0) ++ a;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;