Theorem hbov 101

Description: Hypothesis builder for binary operation.

Hypotheses

Ref	Expression
hbov.1	|- F:(be -> (ga -> de))
hbov.2	|- A:be
hbov.3	|- B:al
hbov.4	|- C:ga
hbov.5	|- R |= [(\x:al FB) = F]
hbov.6	|- R |= [(\x:al AB) = A]
hbov.7	|- R |= [(\x:al CB) = C]

Assertion

Ref	Expression
hbov	|- R |= [(\x:al [AFC]B) = [AFC]]

Proof of Theorem hbov

Step	Hyp	Ref	Expression
1		hbov.5	. . . 4 |- R |= [(\x:al FB) = F]
2	1	ax-cb1 29	. . 3 |- R:*
3	2	trud 27	. 2 |- R |= T.
4		hbov.1	. . . 4 |- F:(be -> (ga -> de))
5		hbov.2	. . . 4 |- A:be
6		hbov.4	. . . 4 |- C:ga
7	4, 5, 6	wov 64	. . 3 |- [AFC]:de
8		hbov.3	. . 3 |- B:al
9		weq 38	. . . 4 |- = :(de -> (de -> *))
10	4, 5	wc 45	. . . . 5 |- (FA):(ga -> de)
11	10, 6	wc 45	. . . 4 |- ((FA)C):de
12	4, 5, 6	df-ov 65	. . . 4 |- T. |= (( = [AFC])((FA)C))
13	9, 7, 11, 12	dfov2 67	. . 3 |- T. |= [[AFC] = ((FA)C)]
14		hbov.6	. . . . . 6 |- R |= [(\x:al AB) = A]
15	4, 5, 8, 1, 14	hbc 100	. . . . 5 |- R |= [(\x:al (FA)B) = (FA)]
16		hbov.7	. . . . 5 |- R |= [(\x:al CB) = C]
17	10, 6, 8, 15, 16	hbc 100	. . . 4 |- R |= [(\x:al ((FA)C)B) = ((FA)C)]
18		wtru 40	. . . 4 |- T.:*
19	17, 18	adantr 50	. . 3 |- (R, T.) |= [(\x:al ((FA)C)B) = ((FA)C)]
20	7, 8, 13, 19	hbxfrf 97	. 2 |- (R, T.) |= [(\x:al [AFC]B) = [AFC]]
21	3, 20	mpdan 33	1 |- R |= [(\x:al [AFC]B) = [AFC]]

Syntax hints:   -> ht 2  kc 5  \kl 6   = ke 7  T.kt 8  [kbr 9   |= wffMMJ2 11  wffMMJ2t 12

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-distrc 61  ax-leq 62

This theorem depends on definitions:  df-ov 65

This theorem is referenced by:  clf  105  hbct  145  exlimdv  157  cbvf  167  leqf  169  exlimd  171  exmid  186  axrep  207