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Proof step 7 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 \ (bag ++ a0)) = (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ 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)

Rule argument

Additional information

Used simplifier rules: diff-rec: ⊦ bag0 \ (bag1 ++ a) = (bag0 -- a) \ bag1;