Theorem ddif 3635

Description: Double complement under universal class. Exercise 4.10(s) of [Mendelson] p. 231. (Contributed by NM, 8-Jan-2002.)

Assertion

Ref	Expression
ddif

Proof of Theorem ddif

Dummy variable is distinct from all other variables.

Step	Hyp	Ref	Expression
1		vex 3112	. . . . 5
2		eldif 3485	. . . . 5
3	1, 2	mpbiran 918	. . . 4
4	3	con2bii 332	. . 3
5	1	biantrur 506	. . 3
6	4, 5	bitr2i 250	. 2
7	6	difeqri 3623	1

Syntax hints:  -.wn 3  /\wa 369  =wceq 1395  e.wcel 1818  cvv 3109  \cdif 3472

This theorem is referenced by:  dfun3  3735  dfin3  3736  invdif  3738  ssindif0  3880  difdifdir  3915

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1618  ax-4 1631  ax-5 1704  ax-6 1747  ax-7 1790  ax-10 1837  ax-11 1842  ax-12 1854  ax-13 1999  ax-ext 2435

This theorem depends on definitions:  df-bi 185  df-an 371  df-tru 1398  df-ex 1613  df-nf 1617  df-sb 1740  df-clab 2443  df-cleq 2449  df-clel 2452  df-nfc 2607  df-v 3111  df-dif 3478