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Proof step 9 from proof fidb-diff-one (previous step)

count(fid0, fidb \ {fid1} ) = 0 ⊦ count(fid0, fidb) = 0, fid0 = fid1

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: bag-union, inst: fidbag name: difference seq: ⊦ count(fid, fidb \ fidb0) = (count(fid, fidb0) ≤ count(fid, fidb) ⊃ count(fid, fidb) - count(fid, fidb0) ;0 ) subst: [fidb0, fidb, fid] → [{fid1} , fidb, fid0] flags: <all paths>

Additional information

Used simplifier rules: ⊦ count(fid, {fid0} ) = (fid = fid0 ⊃ 1;0); (s)% ⊦ 0 ≤ m; (s)% ⊦ m - 0 = m; (s) Used terms: {fid1} , fidb, fid0