Theorem ceq12 78

Description: Equality theorem for combination.

Hypotheses

Ref	Expression
ceq12.1	|- F:(al -> be)
ceq12.2	|- A:al
ceq12.3	|- R |= [F = T]
ceq12.4	|- R |= [A = B]

Assertion

Ref	Expression
ceq12	|- R |= [(FA) = (TB)]

Proof of Theorem ceq12

Step	Hyp	Ref	Expression
1		weq 38	. 2 |- = :(be -> (be -> *))
2		ceq12.1	. . 3 |- F:(al -> be)
3		ceq12.2	. . 3 |- A:al
4	2, 3	wc 45	. 2 |- (FA):be
5		ceq12.3	. . . 4 |- R |= [F = T]
6	2, 5	eqtypi 69	. . 3 |- T:(al -> be)
7		ceq12.4	. . . 4 |- R |= [A = B]
8	3, 7	eqtypi 69	. . 3 |- B:al
9	6, 8	wc 45	. 2 |- (TB):be
10		weq 38	. . . 4 |- = :((al -> be) -> ((al -> be) -> *))
11	10, 2, 6, 5	dfov1 66	. . 3 |- R |= (( = F)T)
12		weq 38	. . . 4 |- = :(al -> (al -> *))
13	12, 3, 8, 7	dfov1 66	. . 3 |- R |= (( = A)B)
14	2, 6, 3, 8	ax-ceq 46	. . 3 |- ((( = F)T), (( = A)B)) |= (( = (FA))(TB))
15	11, 13, 14	syl2anc 19	. 2 |- R |= (( = (FA))(TB))
16	1, 4, 9, 15	dfov2 67	1 |- R |= [(FA) = (TB)]

Syntax hints:   -> ht 2  kc 5   = 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

This theorem depends on definitions:  df-ov 65

This theorem is referenced by:  ceq1  79  ceq2  80  oveq123  88  hbc  100  ac  184