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Proof step 10 from proof difference (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ a ≠ b → count(a, bag ++ b) = count(a, bag)
Proof step 12:
¬ a ≠ a0, count(a, (bag0 -- a0) \ bag) = (count(a, bag) < count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) ⊃ count(a, bag0) - (a = a0 ∧ count(a0, bag0) ≠ 0 ⊃ 1;0) - count(a, bag);0) ⊦ count(a, (bag0 -- a0) \ bag) = (count(a, bag ++ a0) < count(a, bag0) ⊃ count(a, bag0) - count(a, bag ++ a0);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, bag, a] flags: rewrite

Additional information

Used simplifier rules: sub-base: ⊦ m - 0 = m; from specification nat Used terms: a0, bag, a