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Proof step 21 from proof bag-rem-empty (previous step)

count(n, bag) = 1, count(n1, bag -- n) = 0, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 22:
⊦ count(n, bag0 -- n0) = count(n, bag0) - (n = n0 ∧ count(n, bag0) ≠ 0 ⊃ 1;0)
Proof step 23:
count(n1, bag) - (n1 = n ∧ count(n1, bag) ≠ 0 ⊃ 1;0) = 0, count(n, bag) = 1, ∀ n2. count(n2, bag -- n) = 0 ⊦ count(n1, bag) = count(n1, ∅ ++ n)

Rule argument

spec: bag-union, inst: nat-bag name: count-del seq: ⊦ count(n, bag0 -- n0) = count(n, bag0) - (n = n0 ∧ count(n, bag0) ≠ 0 ⊃ 1;0) subst: [n0, bag0, n] → [n, bag, n1] flags: <all paths>

Additional information

Used terms: n, bag, n1