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Proof step 9 from proof bag-add-rem-same (previous step)

∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0 ⊦ (∀ n0. count(n0, (m = n ⊃ bag0;(bag0 -- n) ++ m) ++ n) = count(n0, bag0 ++ m)) ↔ count(n, bag0 ++ m) ≠ 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ count(n, bag0 ++ n0) = (n = n0 ⊃ 1;0) + count(n, bag0)
Proof step 11:
∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0 ⊦ (∀ n0. count(n0, (m = n ⊃ bag0;(bag0 -- n) ++ m) ++ n) = count(n0, bag0 ++ m)) ↔ (n = m ⊃ 1;0) + count(n, bag0) ≠ 0

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, bag0, n] flags: <all paths>

Additional information

Used terms: m, bag0, n