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Proof step 12 from proof bag-rem-empty (previous step)

∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n), count(n0, bag) = 0

User applied rule all left

And obtained the following new goals

Proof step 13:
count(n0, bag -- n) = 0, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n0, bag) = 0, count(n1, bag) = count(n1, ∅ ++ n)

Rule argument

pos: left 1 terms: n0 abort: F discard: F computed: F

Additional information

Used terms: n0