Step
*
3
of Lemma
equal-count-bag-to-set
1. T : Type
2. eq : EqDecider(T)
3. ∀[x:T]. ∀[bs:bag(T)]. uiff(x ↓∈ bs;1 ≤ (#x in bs))
4. ∀[bs:bag(T)]. ∀[x:T]. uiff(x ↓∈ bs;x ↓∈ bag-to-set(eq;bs))
5. bs : bag(T)
6. x : T
7. y : T
8. x ↓∈ bs
⇒ y ↓∈ bs
9. x ↓∈ bs
⇐ y ↓∈ bs
⊢ (#x in bag-to-set(eq;bs)) = (#y in bag-to-set(eq;bs)) ∈ ℤ
BY
{ ((RWO "count-bag-to-set" 0 THEN Auto) THEN RepeatFor 2 (AutoSplit)) }
1
1. T : Type
2. eq : EqDecider(T)
3. ∀[x:T]. ∀[bs:bag(T)]. uiff(x ↓∈ bs;1 ≤ (#x in bs))
4. ∀[bs:bag(T)]. ∀[x:T]. uiff(x ↓∈ bs;x ↓∈ bag-to-set(eq;bs))
5. bs : bag(T)
6. x : T
7. y : T
8. ¬0 < (#y in bs)
9. x ↓∈ bs
⇒ y ↓∈ bs
10. x ↓∈ bs
⇐ y ↓∈ bs
11. 0 < (#x in bs)
⊢ 1 = 0 ∈ ℤ
2
1. T : Type
2. eq : EqDecider(T)
3. ∀[x:T]. ∀[bs:bag(T)]. uiff(x ↓∈ bs;1 ≤ (#x in bs))
4. ∀[bs:bag(T)]. ∀[x:T]. uiff(x ↓∈ bs;x ↓∈ bag-to-set(eq;bs))
5. bs : bag(T)
6. x : T
7. ¬0 < (#x in bs)
8. y : T
9. x ↓∈ bs
⇒ y ↓∈ bs
10. x ↓∈ bs
⇐ y ↓∈ bs
11. 0 < (#y in bs)
⊢ 0 = 1 ∈ ℤ
Latex:
Latex:
1. T : Type
2. eq : EqDecider(T)
3. \mforall{}[x:T]. \mforall{}[bs:bag(T)]. uiff(x \mdownarrow{}\mmember{} bs;1 \mleq{} (\#x in bs))
4. \mforall{}[bs:bag(T)]. \mforall{}[x:T]. uiff(x \mdownarrow{}\mmember{} bs;x \mdownarrow{}\mmember{} bag-to-set(eq;bs))
5. bs : bag(T)
6. x : T
7. y : T
8. x \mdownarrow{}\mmember{} bs {}\mRightarrow{} y \mdownarrow{}\mmember{} bs
9. x \mdownarrow{}\mmember{} bs \mLeftarrow{}{} y \mdownarrow{}\mmember{} bs
\mvdash{} (\#x in bag-to-set(eq;bs)) = (\#y in bag-to-set(eq;bs))
By
Latex:
((RWO "count-bag-to-set" 0 THEN Auto) THEN RepeatFor 2 (AutoSplit))
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