Theorem symdifeq2 3250

Description: Equality law for intersection. (Contributed by SF, 11-Jan-2015.)

Assertion

Ref	Expression
symdifeq2	|- (A = B -> (C (+) A) = (C (+) B))

Proof of Theorem symdifeq2

Step	Hyp	Ref	Expression
1		difeq2 3248	. . . 4 |- (A = B -> (C \ A) = (C \ B))
2	1	compleqd 3246	. . 3 |- (A = B -> ∼ (C \ A) = ∼ (C \ B))
3		difeq1 3247	. . . 4 |- (A = B -> (A \ C) = (B \ C))
4	3	compleqd 3246	. . 3 |- (A = B -> ∼ (A \ C) = ∼ (B \ C))
5	2, 4	nineq12d 3243	. 2 |- (A = B -> ( ∼ (C \ A) &ncap ∼ (A \ C)) = ( ∼ (C \ B) &ncap ∼ (B \ C)))
6		df-symdif 3217	. . 3 |- (C (+) A) = ((C \ A) u. (A \ C))
7		df-un 3215	. . 3 |- ((C \ A) u. (A \ C)) = ( ∼ (C \ A) &ncap ∼ (A \ C))
8	6, 7	eqtri 2373	. 2 |- (C (+) A) = ( ∼ (C \ A) &ncap ∼ (A \ C))
9		df-symdif 3217	. . 3 |- (C (+) B) = ((C \ B) u. (B \ C))
10		df-un 3215	. . 3 |- ((C \ B) u. (B \ C)) = ( ∼ (C \ B) &ncap ∼ (B \ C))
11	9, 10	eqtri 2373	. 2 |- (C (+) B) = ( ∼ (C \ B) &ncap ∼ (B \ C))
12	5, 8, 11	3eqtr4g 2410	1 |- (A = B -> (C (+) A) = (C (+) B))

Syntax hints:   -> wi 4   = wceq 1642   &ncap cnin 3205   ∼ ccompl 3206   \ cdif 3207   u. cun 3208   (+) csymdif 3210

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-in 3214  df-un 3215  df-dif 3216  df-symdif 3217

This theorem is referenced by:  symdifeq12  3251  symdifeq2i  3253  symdifeq2d  3256