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Proof step 4 from proof bag-diff-empty (previous step)

⊦ count(n, bag \ bag) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ count(n, bag \ bag2) = (count(n, bag2) ≤ count(n, bag) ⊃ count(n, bag) - count(n, bag2);0)

Rule argument

spec: bag-union, inst: nat-bag name: difference seq: ⊦ count(n, bag \ bag2) = (count(n, bag2) ≤ count(n, bag) ⊃ count(n, bag) - count(n, bag2) ;0 ) subst: [bag2, bag, n] → [bag, bag, n] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ m; (s)% ⊦ m - m = 0; (s) Used terms: bag, bag, n