Theorem hbct 145

Description: Hypothesis builder for context conjunction.

Hypotheses

Ref	Expression
hbct.1	|- A:*
hbct.2	|- B:al
hbct.3	|- C:*
hbct.4	|- R |= [(\x:al AB) = A]
hbct.5	|- R |= [(\x:al CB) = C]

Assertion

Ref	Expression
hbct	|- R |= [(\x:al (A, C)B) = (A, C)]

Proof of Theorem hbct

Step	Hyp	Ref	Expression
1		hbct.4	. . . 4 |- R |= [(\x:al AB) = A]
2	1	ax-cb1 29	. . 3 |- R:*
3	2	trud 27	. 2 |- R |= T.
4		hbct.1	. . . 4 |- A:*
5		hbct.3	. . . 4 |- C:*
6	4, 5	wct 44	. . 3 |- (A, C):*
7		hbct.2	. . 3 |- B:al
8		wan 126	. . . . 5 |- /\ :(* -> (* -> *))
9	8, 4, 5	wov 64	. . . 4 |- [A /\ C]:*
10	4, 5	dfan2 144	. . . 4 |- T. |= [[A /\ C] = (A, C)]
11	9, 10	eqcomi 70	. . 3 |- T. |= [(A, C) = [A /\ C]]
12	8, 7, 2	a17i 96	. . . . 5 |- R |= [(\x:al /\ B) = /\ ]
13		hbct.5	. . . . 5 |- R |= [(\x:al CB) = C]
14	8, 4, 7, 5, 12, 1, 13	hbov 101	. . . 4 |- R |= [(\x:al [A /\ C]B) = [A /\ C]]
15		wtru 40	. . . 4 |- T.:*
16	14, 15	adantr 50	. . 3 |- (R, T.) |= [(\x:al [A /\ C]B) = [A /\ C]]
17	6, 7, 11, 16	hbxfrf 97	. 2 |- (R, T.) |= [(\x:al (A, C)B) = (A, C)]
18	3, 17	mpdan 33	1 |- R |= [(\x:al (A, C)B) = (A, C)]

Syntax hints:   -> ht 2  *hb 3  kc 5  \kl 6   = ke 7  T.kt 8  [kbr 9  kct 10   |= wffMMJ2 11  wffMMJ2t 12   /\ tan 109

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-ded 43  ax-ceq 46  ax-beta 60  ax-distrc 61  ax-leq 62  ax-distrl 63  ax-hbl1 93  ax-17 95  ax-inst 103

This theorem depends on definitions:  df-ov 65  df-an 118

This theorem is referenced by:  alimdv  172  ax5  194