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Proof step 4 from proof del-size (previous step)

⊦ #(bag -- a) = # bag - 1, count(a, bag) = 0

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ a ∈ bag → (∃ bag0. bag = bag0 ++ a)

Rule argument

spec: inst: name: exin seq: ⊦ a ∈ bag → (∃ bag0. bag = bag0 ++ a) subst: [a, bag] → [a, bag] flags: current, []

Additional information

Used simplifier rules: in: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0;%del-rec: ⊦ (bag ++ a) -- b = (a = b ⊃ bag;(bag -- b) ++ a);%size-rec: ⊦ #(bag ++ a) = # bag + 1;%count-same: ⊦ count(a, bag ++ a) = count(a, bag) + 1; from specification bag-basic%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2 Used terms: a, bag