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Proof step 5 from proof diff-add-sw (previous step)

⊦ (count(a0, bag1) ≤ count(a0, bag0 ++ a) ⊃ count(a0, bag0 ++ a) - count(a0, bag1);0) = count(a0, (bag0 \ bag1) ++ a), a ∈ bag1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ∈ bag ↔ count(a, bag) ≠ 0
Proof step 7:
⊦ (count(a0, bag1) ≤ count(a0, bag0 ++ a) ⊃ count(a0, bag0 ++ a) - count(a0, bag1);0) = count(a0, (bag0 \ bag1) ++ a), count(a, bag1) ≠ 0

Rule argument

spec: inst: name: in seq: ⊦ a ∈ bag ↔ count(a, bag) ≠ 0 subst: [bag, a] → [bag1, a] flags: rewrite, current

Additional information

Used terms: bag1, a