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

⊦ #((s ∪ s0) ++ a) = #(s ∪ s0) + 1, a ∈ s0, a ∈ s

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # s + 1 = #(s ++ a) ↔ ¬ a ∈ s; (s)