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Proof step 20 from proof union-size-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
a ∈ s, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ s = ∅, #(s ∪ s0) = # s0 + # s - #(s0 ∩ s), # s = 0

Rule argument

Additional information

Used simplifier rules: uninsert: ⊦ a ∈ s → (s ∪ s0) ++ a = s ∪ s0;%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