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Proof step 13 from proof union-size-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
a ∈ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ s0 = ∅, #(s ∪ s0) = # s0 + # s + 1 - #(s0 ∩ (s ++ a)), # s0 = 0, a ∈ s

Rule argument

Additional information

Used simplifier rules: unrax: ⊦ a ∈ s ∪ s0 ↔ ¬ ¬(a ∈ s ∨ a ∈ 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