Theorem eqcomx 52

Description: Commutativity of equality. (Contributed by Mario Carneiro, 7-Oct-2014.)

Hypotheses

Ref	Expression
eqcomx.1	|- A:al
eqcomx.2	|- B:al
eqcomx.3	|- R |= (( = A)B)

Assertion

Ref	Expression
eqcomx	|- R |= (( = B)A)

Proof of Theorem eqcomx

Step	Hyp	Ref	Expression
1		eqcomx.3	. . . 4 |- R |= (( = A)B)
2	1	ax-cb1 29	. . 3 |- R:*
3		eqcomx.1	. . . 4 |- A:al
4	3	ax-refl 42	. . 3 |- T. |= (( = A)A)
5	2, 4	a1i 28	. 2 |- R |= (( = A)A)
6		weq 41	. . . . . 6 |- = :(al -> (al -> *))
7	6	ax-refl 42	. . . . 5 |- T. |= (( = = ) = )
8	2, 7	a1i 28	. . . 4 |- R |= (( = = ) = )
9		eqcomx.2	. . . . 5 |- B:al
10	6, 6, 3, 9	ax-ceq 51	. . . 4 |- ((( = = ) = ), (( = A)B)) |= (( = ( = A))( = B))
11	8, 1, 10	syl2anc 19	. . 3 |- R |= (( = ( = A))( = B))
12	6, 3	wc 50	. . . 4 |- ( = A):(al -> *)
13	6, 9	wc 50	. . . 4 |- ( = B):(al -> *)
14	12, 13, 3, 3	ax-ceq 51	. . 3 |- ((( = ( = A))( = B)), (( = A)A)) |= (( = (( = A)A))(( = B)A))
15	11, 5, 14	syl2anc 19	. 2 |- R |= (( = (( = A)A))(( = B)A))
16	5, 15	ax-eqmp 45	1 |- R |= (( = B)A)

Syntax hints:   -> ht 2  *hb 3  kc 5   = ke 7   |= wffMMJ2 11  wffMMJ2t 12

This theorem was proved from axioms:  ax-syl 15  ax-jca 17  ax-trud 26  ax-cb1 29  ax-weq 40  ax-refl 42  ax-eqmp 45  ax-wc 49  ax-ceq 51

This theorem is referenced by:  mpbirx  53  eqcomi  79