Theorem hbl 102

Description: Hypothesis builder for lambda abstraction.

Hypotheses

Ref	Expression
hbl.1	|- A:ga
hbl.2	|- B:al
hbl.3	|- R |= [(\x:al AB) = A]

Assertion

Ref	Expression
hbl	|- R |= [(\x:al \y:be AB) = \y:be A]

Distinct variable groups:   x,y   y,B   y,R

Proof of Theorem hbl

Step	Hyp	Ref	Expression
1		hbl.1	. . . . 5 |- A:ga
2	1	wl 59	. . . 4 |- \y:be A:(be -> ga)
3	2	wl 59	. . 3 |- \x:al \y:be A:(al -> (be -> ga))
4		hbl.2	. . 3 |- B:al
5	3, 4	wc 45	. 2 |- (\x:al \y:be AB):(be -> ga)
6		hbl.3	. . . 4 |- R |= [(\x:al AB) = A]
7	6	ax-cb1 29	. . 3 |- R:*
8	1, 4	distrl 84	. . 3 |- T. |= [(\x:al \y:be AB) = \y:be (\x:al AB)]
9	7, 8	a1i 28	. 2 |- R |= [(\x:al \y:be AB) = \y:be (\x:al AB)]
10	1	wl 59	. . . 4 |- \x:al A:(al -> ga)
11	10, 4	wc 45	. . 3 |- (\x:al AB):ga
12	11, 6	leq 81	. 2 |- R |= [\y:be (\x:al AB) = \y:be A]
13	5, 9, 12	eqtri 85	1 |- R |= [(\x:al \y:be AB) = \y:be A]

Syntax hints:   -> ht 2  kc 5  \kl 6   = ke 7  [kbr 9   |= wffMMJ2 11  wffMMJ2t 12

This theorem was proved from axioms:  ax-syl 15  ax-jca 17  ax-trud 26  ax-cb1 29  ax-cb2 30  ax-refl 39  ax-eqmp 42  ax-ceq 46  ax-leq 62  ax-distrl 63

This theorem depends on definitions:  df-ov 65

This theorem is referenced by:  cbvf  167  ax7  196  axrep  207