Theorem ceq1 79

Description: Equality theorem for combination.

Hypotheses

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

Assertion

Ref	Expression
ceq1	|- R |= [(FA) = (TA)]

Proof of Theorem ceq1

Step	Hyp	Ref	Expression
1		ceq12.1	. 2 |- F:(al -> be)
2		ceq12.2	. 2 |- A:al
3		ceq12.3	. 2 |- R |= [F = T]
4	3	ax-cb1 29	. . 3 |- R:*
5	4, 2	eqid 73	. 2 |- R |= [A = A]
6	1, 2, 3, 5	ceq12 78	1 |- R |= [(FA) = (TA)]

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:  hbxfrf  97  ovl  107  alval  132  exval  133  euval  134  notval  135  ax4g  139  dfan2  144  eta  166  ac  184  ax14  204  axrep  207  axpow  208  axun  209