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Proof step 6 from proof union-sizele (previous step)

#(s ∪ s0) ≤ # s ⊦ #(s ∪ s0) = # s, ∀ a. a ∈ s0 → a ∈ s

User applied rule cut formula

And obtained the following new goals

Proof step 7:
#(s ∪ s0) ≤ # s ⊦ #(s ∪ s0) < # s, #(s ∪ s0) = # s, ∀ a. a ∈ s0 → a ∈ s

Rule argument

fma: #(s ∪ s0) < # s

Additional information

Used simplifier rules: union-sizeeq: ⊦ #(s ∪ s0) = # s ↔ s0 ⊆ s;%una: ⊦ (s ∪ s0) ∪ s1 = s ∪ s0 ∪ s1;%unc: ⊦ s ∪ s0 = s0 ∪ s;%union-size: ⊦ ¬ #(s ∪ s0) < # s;%Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2); from specification set-basic Used formulas: #(s ∪ s0) < # s