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Proof step 3 from proof diff-size (
previous step
)
bag ⊆ bag1, ∀ bag0. bag ⊆ bag0 → #(bag0 \ bag) = # bag0 - # bag ⊦ #(bag1 \ bag) = # bag1 - # bag
User applied rule all left
And closed the goal!
Rule argument
pos: left 2 terms: bag1 abort: F discard: T computed: T
Additional information
Used terms: bag1