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Proof step 3 from proof fidb-diff-one (previous step)

fid0 ∈ fidb ⊦ fid0 = fid1, fid0 ∈ fidb \ {fid1}

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ fid ∈ fidb ↔ count(fid, fidb) ≠ 0
Proof step 5:
⊦ count(fid0, fidb) = 0, fid0 = fid1, fid0 ∈ fidb \ {fid1}

Rule argument

spec: bag-union, inst: fidbag name: in seq: ⊦ fid ∈ fidb ↔ count(fid, fidb) ≠ 0 subst: [fidb, fid] → [fidb, fid0] flags: <all paths>

Additional information

Used terms: fidb, fid0