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Proof step 4 from proof union-size-02 (previous step)

a0 ∈ s0, a ∈ s0, a ∈ s ∪ s0 ⊦ #(s ∪ s0) = #((s ∪ s0) -- a) + 1, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ s0, a ∈ s0 ⊦ #(s ∪ s0) = (#(s ∪ s0) - 1) + 1, a ∈ s

Rule argument

Additional information

Used simplifier rules: del-size: ⊦ a ∈ s → #(s -- a) = # s - 1;%Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;