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Proof step 11 from proof subset-plusplus (previous step)

count(a0, bag1 ++ a) < count(a0, bag0 ++ a), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a0, bag0 ++ a) = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)
Proof step 13:
¬ a0 ≠ a, count(a0, bag1 ++ a) < count(a0, bag0 ++ a), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a0, bag0 ++ a) = 0
Proof step 15:
a0 ≠ a, count(a0, bag1) < count(a0, bag0 ++ a), ∀ a. count(a, bag0) ≤ count(a, bag1) ⊦ count(a0, bag0 ++ a) = 0

Rule argument

spec: bag-basic inst: name: count-other seq: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag) subst: [b, bag, a] → [a, bag1, a0] flags: rewrite

Additional information

Used terms: a, bag1, a0