Step
*
1
1
1
3
of Lemma
bag-parts'-no-repeats
1. T : Type
2. valueall-type(T)
3. eq : EqDecider(T)
4. x : T
5. bs : bag(T)
6. ¬(bs = {} ∈ bag(T))
7. bag-no-repeats(bag(T) List+;bag-map(λL.[{} / L];bag-parts(eq;bs)))
8. bag-no-repeats(bag(T) List+;[L∈bag-parts(eq;bs)|((#x in hd(L)) =z 0)])
9. z : bag(T) List+
10. z ↓∈ bag-map(λL.[{} / L];bag-parts(eq;bs))
⊢ ¬z ↓∈ [L∈bag-parts(eq;bs)|((#x in hd(L)) =z 0)]
BY
{ xxx((D 0 THENA Auto) THEN Try ((D -1 THEN Complete (Auto))))xxx }
1
1. T : Type
2. valueall-type(T)
3. eq : EqDecider(T)
4. x : T
5. bs : bag(T)
6. ¬(bs = {} ∈ bag(T))
7. bag-no-repeats(bag(T) List+;bag-map(λL.[{} / L];bag-parts(eq;bs)))
8. bag-no-repeats(bag(T) List+;[L∈bag-parts(eq;bs)|((#x in hd(L)) =z 0)])
9. z : bag(T) List+
10. z ↓∈ bag-map(λL.[{} / L];bag-parts(eq;bs))
11. z ↓∈ [L∈bag-parts(eq;bs)|((#x in hd(L)) =z 0)]
⊢ False
Latex:
Latex:
1. T : Type
2. valueall-type(T)
3. eq : EqDecider(T)
4. x : T
5. bs : bag(T)
6. \mneg{}(bs = \{\})
7. bag-no-repeats(bag(T) List\msupplus{};bag-map(\mlambda{}L.[\{\} / L];bag-parts(eq;bs)))
8. bag-no-repeats(bag(T) List\msupplus{};[L\mmember{}bag-parts(eq;bs)|((\#x in hd(L)) =\msubz{} 0)])
9. z : bag(T) List\msupplus{}
10. z \mdownarrow{}\mmember{} bag-map(\mlambda{}L.[\{\} / L];bag-parts(eq;bs))
\mvdash{} \mneg{}z \mdownarrow{}\mmember{} [L\mmember{}bag-parts(eq;bs)|((\#x in hd(L)) =\msubz{} 0)]
By
Latex:
xxx((D 0 THENA Auto) THEN Try ((D -1 THEN Complete (Auto))))xxx
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