Theorem alval 142

Description: Value of the for all predicate. (Contributed by Mario Carneiro, 8-Oct-2014.)

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 134	. . 3 |- A.:((al -> *) -> *)
2		alval.1	. . 3 |- F:(al -> *)
3	1, 2	wc 50	. 2 |- (A.F):*
4		df-al 126	. . 3 |- T. |= [A. = \p:(al -> *) [p:(al -> *) = \x:al T.]]
5	1, 2, 4	ceq1 89	. 2 |- T. |= [(A.F) = (\p:(al -> *) [p:(al -> *) = \x:al T.]F)]
6		wv 64	. . . 4 |- p:(al -> *):(al -> *)
7		wtru 43	. . . . 5 |- T.:*
8	7	wl 66	. . . 4 |- \x:al T.:(al -> *)
9	6, 8	weqi 76	. . 3 |- [p:(al -> *) = \x:al T.]:*
10		weq 41	. . . 4 |- = :((al -> *) -> ((al -> *) -> *))
11	6, 2	weqi 76	. . . . 5 |- [p:(al -> *) = F]:*
12	11	id 25	. . . 4 |- [p:(al -> *) = F] |= [p:(al -> *) = F]
13	10, 6, 8, 12	oveq1 99	. . 3 |- [p:(al -> *) = F] |= [[p:(al -> *) = \x:al T.] = [F = \x:al T.]]
14	9, 2, 13	cl 116	. 2 |- T. |= [(\p:(al -> *) [p:(al -> *) = \x:al T.]F) = [F = \x:al T.]]
15	3, 5, 14	eqtri 95	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 122

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-wctl 31  ax-wctr 32  ax-weq 40  ax-refl 42  ax-eqmp 45  ax-wc 49  ax-ceq 51  ax-wv 63  ax-wl 65  ax-beta 67  ax-distrc 68  ax-leq 69  ax-wov 71  ax-eqtypi 77  ax-eqtypri 80  ax-hbl1 103  ax-17 105  ax-inst 113

This theorem depends on definitions:  df-ov 73  df-al 126

This theorem is referenced by:  ax4g  149  alrimiv  151  olc  164  orc  165  alrimi  182