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Proof step 8 from proof difference (previous step)

∀ a, bag0. count(a, bag0 \ bag) = (count(a, bag) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag);0) ⊦ count(a, (bag0 -- a0) \ bag) = (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0)

User applied rule all left

And obtained the following new goals

Proof step 9:
count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0 -- a0) ⊃ count(a, bag0 -- a0) - count(a, bag);0) ⊦ count(a, (bag0 -- a0) \ bag) = (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0)

Rule argument

pos: left 1 terms: a, bag0 -- a0 abort: F discard: T computed: T

Additional information

Used terms: a, bag0 -- a0