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Proof step 4 from proof bag-add-rev (previous step)

⊦ count(n0, bag ++ n ++ m) = count(n0, bag ++ m ++ n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ count(n, bag0 ++ n0) = (n = n0 ⊃ 1;0) + count(n, bag0)
Proof step 6:
⊦ (n0 = m ⊃ 1;0) + count(n0, bag ++ n) = count(n0, bag ++ m ++ n)

Rule argument

spec: bag-basic, inst: nat-bag name: count-insert seq: ⊦ count(n, bag0 ++ n0) = (n = n0 ⊃ 1;0) + count(n, bag0) subst: [n0, bag0, n] → [m, bag ++ n, n0] flags: <all paths>

Additional information

Used terms: m, bag ++ n, n0