Theorem alval 142

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

Hypothesis

Ref	Expression
alval.1	⊢ F:(α → ∗)

Assertion

Ref	Expression
alval	⊢ ⊤⊧[(∀F) = [F = λx:α ⊤]]

Distinct variable group:   α,x

Proof of Theorem alval

Dummy variable p is distinct from all other variables.

Step	Hyp	Ref	Expression
1		wal 134	. . 3 ⊢ ∀:((α → ∗) → ∗)
2		alval.1	. . 3 ⊢ F:(α → ∗)
3	1, 2	wc 50	. 2 ⊢ (∀F):∗
4		df-al 126	. . 3 ⊢ ⊤⊧[∀ = λp:(α → ∗) [p:(α → ∗) = λx:α ⊤]]
5	1, 2, 4	ceq1 89	. 2 ⊢ ⊤⊧[(∀F) = (λp:(α → ∗) [p:(α → ∗) = λx:α ⊤]F)]
6		wv 64	. . . 4 ⊢ p:(α → ∗):(α → ∗)
7		wtru 43	. . . . 5 ⊢ ⊤:∗
8	7	wl 66	. . . 4 ⊢ λx:α ⊤:(α → ∗)
9	6, 8	weqi 76	. . 3 ⊢ [p:(α → ∗) = λx:α ⊤]:∗
10		weq 41	. . . 4 ⊢ = :((α → ∗) → ((α → ∗) → ∗))
11	6, 2	weqi 76	. . . . 5 ⊢ [p:(α → ∗) = F]:∗
12	11	id 25	. . . 4 ⊢ [p:(α → ∗) = F]⊧[p:(α → ∗) = F]
13	10, 6, 8, 12	oveq1 99	. . 3 ⊢ [p:(α → ∗) = F]⊧[[p:(α → ∗) = λx:α ⊤] = [F = λx:α ⊤]]
14	9, 2, 13	cl 116	. 2 ⊢ ⊤⊧[(λp:(α → ∗) [p:(α → ∗) = λx:α ⊤]F) = [F = λx:α ⊤]]
15	3, 5, 14	eqtri 95	1 ⊢ ⊤⊧[(∀F) = [F = λx:α ⊤]]

Syntax hints:  tv 1   → ht 2  ∗hb 3  kc 5  λkl 6   = ke 7  ⊤kt 8  [kbr 9  ⊧wffMMJ2 11  wffMMJ2t 12  ∀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