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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.
StepHypRef Expression
1 wal 134 . . 3 |- A.:((al -> *) -> *)
2 alval.1 . . 3 |- F:(al -> *)
31, 2wc 50 . 2 |- (A.F):*
4 df-al 126 . . 3 |- T. |= [A. = \p:(al -> *) [p:(al -> *) = \x:al T.]]
51, 2, 4ceq1 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.:*
87wl 66 . . . 4 |- \x:al T.:(al -> *)
96, 8weqi 76 . . 3 |- [p:(al -> *) = \x:al T.]:*
10 weq 41 . . . 4 |- = :((al -> *) -> ((al -> *) -> *))
116, 2weqi 76 . . . . 5 |- [p:(al -> *) = F]:*
1211id 25 . . . 4 |- [p:(al -> *) = F] |= [p:(al -> *) = F]
1310, 6, 8, 12oveq1 99 . . 3 |- [p:(al -> *) = F] |= [[p:(al -> *) = \x:al T.] = [F = \x:al T.]]
149, 2, 13cl 116 . 2 |- T. |= [(\p:(al -> *) [p:(al -> *) = \x:al T.]F) = [F = \x:al T.]]
153, 5, 14eqtri 95 1 |- T. |= [(A.F) = [F = \x:al T.]]
Colors of variables: type var term
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
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