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Proof step 5 from proof bag-add-rem-same (previous step)

⊦ (∀ n0. count(n0, (bag -- n) ++ n) = count(n0, bag)) ↔ count(n, bag) ≠ 0

User applied rule structural induction

And obtained the following new goals

Proof step 6:
⊦ (∀ n0. count(n0, (∅ -- n) ++ n) = count(n0, ∅)) ↔ count(n, ∅) ≠ 0
Proof step 8:
∀ n. (∀ n0. count(n0, (bag0 -- n) ++ n) = count(n0, bag0)) ↔ count(n, bag0) ≠ 0 ⊦ (∀ n0. count(n0, ((bag0 ++ m) -- n) ++ n) = count(n0, bag0 ++ m)) ↔ count(n, bag0 ++ m) ≠ 0

Rule argument

indvar: bag

Additional information

Used simplifier rules: nat-bag generated by ∅, ++;;