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Proof step 29 from proof bag-add-rem-same (previous step)

∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0 ⊦ 1 + count(n, bag0 -- n) = count(n, bag0), count(n, bag0) = 0, n = m

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ count(n, bag0 -- n0) = count(n, bag0) - (n = n0 ∧ count(n, bag0) ≠ 0 ⊃ 1;0)

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, bag0, n] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic Used terms: n, bag0, n