Theorem alval 132

Description: Value of the for all predicate.

Hypothesis

Ref	Expression
alval.1	|- F:(al -> *)

Assertion

Ref	Expression
alval	|- T. |= [(A.F) = [F = \x:al T.]]

Distinct variable group:   al,x

Proof of Theorem alval

Dummy variable p is distinct from all other variables.

Step	Hyp	Ref	Expression
1		wal 124	. . 3 |- A.:((al -> *) -> *)
2		alval.1	. . 3 |- F:(al -> *)
3	1, 2	wc 45	. 2 |- (A.F):*
4		df-al 116	. . 3 |- T. |= [A. = \p:(al -> *) [p:(al -> *) = \x:al T.]]
5	1, 2, 4	ceq1 79	. 2 |- T. |= [(A.F) = (\p:(al -> *) [p:(al -> *) = \x:al T.]F)]
6		wv 58	. . . 4 |- p:(al -> *):(al -> *)
7		wtru 40	. . . . 5 |- T.:*
8	7	wl 59	. . . 4 |- \x:al T.:(al -> *)
9	6, 8	weqi 68	. . 3 |- [p:(al -> *) = \x:al T.]:*
10		weq 38	. . . 4 |- = :((al -> *) -> ((al -> *) -> *))
11	6, 2	weqi 68	. . . . 5 |- [p:(al -> *) = F]:*
12	11	id 25	. . . 4 |- [p:(al -> *) = F] |= [p:(al -> *) = F]
13	10, 6, 8, 12	oveq1 89	. . 3 |- [p:(al -> *) = F] |= [[p:(al -> *) = \x:al T.] = [F = \x:al T.]]
14	9, 2, 13	cl 106	. 2 |- T. |= [(\p:(al -> *) [p:(al -> *) = \x:al T.]F) = [F = \x:al T.]]
15	3, 5, 14	eqtri 85	1 |- T. |= [(A.F) = [F = \x:al T.]]

Syntax hints:  tv 1   -> ht 2  *hb 3  kc 5  \kl 6   = ke 7  T.kt 8  [kbr 9   |= wffMMJ2 11  wffMMJ2t 12  A.tal 112

This theorem was proved from axioms:  ax-syl 15  ax-jca 17  ax-simpl 20  ax-simpr 21  ax-id 24  ax-trud 26  ax-cb1 29  ax-cb2 30  ax-refl 39  ax-eqmp 42  ax-ceq 46  ax-beta 60  ax-distrc 61  ax-leq 62  ax-hbl1 93  ax-17 95  ax-inst 103

This theorem depends on definitions:  df-ov 65  df-al 116

This theorem is referenced by:  ax4g  139  alrimiv  141  olc  154  orc  155  alrimi  170