Step
*
of Lemma
bag-sum-count
∀[A,B:Type]. ∀[f:B ⟶ bag(A)]. ∀[eq:EqDecider(A)]. ∀[bb:bag(B)]. ∀[z:A].
(bag-sum(bb;b.(#z in f[b])) ~ bag-sum([b∈bb|1 ≤z (#z in f[b])];b.(#z in f[b])))
BY
{ ((Auto THEN BagD (-2)) THEN Auto THEN Subst ⌜as = bs ∈ bag(B)⌝ 0⋅ THEN Auto) }
1
1. A : Type
2. B : Type
3. f : B ⟶ bag(A)
4. eq : EqDecider(A)
5. z : A
6. as : B List
7. bs : B List
8. permutation(B;as;bs)
⊢ bag-sum(bs;b.(#z in f[b])) = bag-sum([b∈bs|1 ≤z (#z in f[b])];b.(#z in f[b])) ∈ ℤ
Latex:
Latex:
\mforall{}[A,B:Type]. \mforall{}[f:B {}\mrightarrow{} bag(A)]. \mforall{}[eq:EqDecider(A)]. \mforall{}[bb:bag(B)]. \mforall{}[z:A].
(bag-sum(bb;b.(\#z in f[b])) \msim{} bag-sum([b\mmember{}bb|1 \mleq{}z (\#z in f[b])];b.(\#z in f[b])))
By
Latex:
((Auto THEN BagD (-2)) THEN Auto THEN Subst \mkleeneopen{}as = bs\mkleeneclose{} 0\mcdot{} THEN Auto)
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