Step
*
1
1
1
1
1
of Lemma
lnk-decl-compatible-single
1. l : IdLnk
2. d : Id List
3. d1 : tg:{tg:Id| (tg ∈ d)} ─→ Type
4. knd : Knd
5. T : Type
6. (↑isrcv(knd))
⇒ (↑lnk(knd) = l)
⇒ (↑tag(knd) ∈ dom(<d, d1>))
⇒ (T = (d1 tag(knd)) ∈ Type)
7. x : Knd@i
8. y : Id
9. (y ∈ d)
10. x = rcv(l,y) ∈ Knd
11. x = knd ∈ Knd
⊢ (d1 tag(knd)) = T ∈ Type
BY
{ D 6 }
1
.....antecedent.....
1. l : IdLnk
2. d : Id List
3. d1 : tg:{tg:Id| (tg ∈ d)} ─→ Type
4. knd : Knd
5. T : Type
6. x : Knd@i
7. y : Id
8. (y ∈ d)
9. x = rcv(l,y) ∈ Knd
10. x = knd ∈ Knd
⊢ ↑isrcv(knd)
2
1. l : IdLnk
2. d : Id List
3. d1 : tg:{tg:Id| (tg ∈ d)} ─→ Type
4. knd : Knd
5. T : Type
6. x : Knd@i
7. y : Id
8. (y ∈ d)
9. x = rcv(l,y) ∈ Knd
10. x = knd ∈ Knd
11. (↑lnk(knd) = l)
⇒ (↑tag(knd) ∈ dom(<d, d1>))
⇒ (T = (d1 tag(knd)) ∈ Type)
⊢ (d1 tag(knd)) = T ∈ Type
Latex:
1. l : IdLnk
2. d : Id List
3. d1 : tg:\{tg:Id| (tg \mmember{} d)\} {}\mrightarrow{} Type
4. knd : Knd
5. T : Type
6. (\muparrow{}isrcv(knd)) {}\mRightarrow{} (\muparrow{}lnk(knd) = l) {}\mRightarrow{} (\muparrow{}tag(knd) \mmember{} dom(<d, d1>)) {}\mRightarrow{} (T = (d1 tag(knd)))
7. x : Knd@i
8. y : Id
9. (y \mmember{} d)
10. x = rcv(l,y)
11. x = knd
\mvdash{} (d1 tag(knd)) = T
By
D 6
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