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Proof step 5 from proof count-del (previous step)

count(a, bag0 -- b) = count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a, (a0 = b ⊃ bag0;(bag0 -- b) ++ a0)) = count(a, bag0 ++ a0) - (a = b ∧ count(b, bag0 ++ a0) ≠ 0 ⊃ 1;0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)
Proof step 7:
¬ a ≠ a0, count(a, bag0 -- b) = count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a, (a0 = b ⊃ bag0;(bag0 -- b) ++ a0)) = count(a, bag0 ++ a0) - (a = b ∧ count(b, bag0 ++ a0) ≠ 0 ⊃ 1;0)
Proof step 9:
a ≠ a0, count(a, bag0 -- b) = count(a, bag0) - (a = b ∧ count(b, bag0) ≠ 0 ⊃ 1;0) ⊦ count(a, (a0 = b ⊃ bag0;(bag0 -- b) ++ a0)) = count(a, bag0) - (a = b ∧ count(b, bag0 ++ a0) ≠ 0 ⊃ 1;0)

Rule argument

spec: bag-basic inst: name: count-other seq: ⊦ a ≠ b → count(a, bag ++ b) = count(a, bag) subst: [b, bag, a] → [a0, bag0, a] flags: rewrite

Additional information

Used terms: a0, bag0, a