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Proof step 19 from proof union-size-03 (
previous step
)
a ∈ s0, ∀ s0. #(s0 ∪ s) = # s0 + # s - #(s0 ∩ s) ⊦ # s0 = 0, s0 = ∅, #(s ∪ s0) = # s0 + # s - #(s0 ∩ s), a ∈ s
User applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: s0 abort: F discard: T computed: F
Additional information
Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s; Used terms: s0