Theorem hbxfrf 97

Description: Transfer a hypothesis builder to an equivalent expression.

Hypotheses

Ref	Expression
hbxfr.1	|- T:be
hbxfr.2	|- B:al
hbxfrf.3	|- R |= [T = A]
hbxfrf.4	|- (S, R) |= [(\x:al AB) = A]

Assertion

Ref	Expression
hbxfrf	|- (S, R) |= [(\x:al TB) = T]

Distinct variable group:   x,R

Proof of Theorem hbxfrf

Step	Hyp	Ref	Expression
1		hbxfr.1	. . . . 5 |- T:be
2		hbxfrf.3	. . . . 5 |- R |= [T = A]
3	1, 2	eqtypi 69	. . . 4 |- A:be
4	3	wl 59	. . 3 |- \x:al A:(al -> be)
5		hbxfr.2	. . 3 |- B:al
6	4, 5	wc 45	. 2 |- (\x:al AB):be
7		hbxfrf.4	. 2 |- (S, R) |= [(\x:al AB) = A]
8	1	wl 59	. . . 4 |- \x:al T:(al -> be)
9	1, 2	leq 81	. . . 4 |- R |= [\x:al T = \x:al A]
10	8, 5, 9	ceq1 79	. . 3 |- R |= [(\x:al TB) = (\x:al AB)]
11	7	ax-cb1 29	. . . 4 |- (S, R):*
12	11	wctl 31	. . 3 |- S:*
13	10, 12	adantl 51	. 2 |- (S, R) |= [(\x:al TB) = (\x:al AB)]
14	2, 12	adantl 51	. 2 |- (S, R) |= [T = A]
15	6, 7, 13, 14	3eqtr4i 86	1 |- (S, R) |= [(\x:al TB) = T]

Syntax hints:  kc 5  \kl 6   = ke 7  [kbr 9  kct 10   |= wffMMJ2 11  wffMMJ2t 12

This theorem was proved from axioms:  ax-syl 15  ax-jca 17  ax-simpl 20  ax-simpr 21  ax-trud 26  ax-cb1 29  ax-cb2 30  ax-refl 39  ax-eqmp 42  ax-ceq 46  ax-leq 62

This theorem depends on definitions:  df-ov 65

This theorem is referenced by:  hbxfr  98  hbov  101  hbct  145