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Proof step 3 from proof union-size-02 (previous step)

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

Heuristic weak cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a0 ∈ s0, a ∈ s0, a ∈ s ∪ s0 ⊦ #(s ∪ s0) = #((s ∪ s0) -- a) + 1, a ∈ s

Rule argument

fma: a ∈ s ∪ s0

Additional information

Used simplifier rules: del-in: ⊦ ¬ a ∈ s → s -- a = s;%Union: ⊦ a ∈ s1 ∪ s2 ↔ a ∈ s1 ∨ a ∈ s2;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;% ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: a ∈ s ∪ s0