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Proof step 21 from proof union-size-03 (
previous step
)
a0 ∈ s0, a ∈ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ #(s ∪ s0) = # s0 + # s - #(s0 ∩ s), a ∈ s
User applied rule all left
And closed the goal!
Rule argument
pos: left 3 terms: s0 abort: F discard: T computed: F
Additional information
Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s; Used terms: s0