Theorem symdifeq1 3248

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

Assertion

Ref	Expression
symdifeq1	|- (A = B -> (A (+) C) = (B (+) C))

Proof of Theorem symdifeq1

Step	Hyp	Ref	Expression
1		difeq1 3246	. . . 4 |- (A = B -> (A \ C) = (B \ C))
2	1	compleqd 3245	. . 3 |- (A = B -> ∼ (A \ C) = ∼ (B \ C))
3		difeq2 3247	. . . 4 |- (A = B -> (C \ A) = (C \ B))
4	3	compleqd 3245	. . 3 |- (A = B -> ∼ (C \ A) = ∼ (C \ B))
5	2, 4	nineq12d 3242	. 2 |- (A = B -> ( ∼ (A \ C) &ncap ∼ (C \ A)) = ( ∼ (B \ C) &ncap ∼ (C \ B)))
6		df-symdif 3216	. . 3 |- (A (+) C) = ((A \ C) u. (C \ A))
7		df-un 3214	. . 3 |- ((A \ C) u. (C \ A)) = ( ∼ (A \ C) &ncap ∼ (C \ A))
8	6, 7	eqtri 2373	. 2 |- (A (+) C) = ( ∼ (A \ C) &ncap ∼ (C \ A))
9		df-symdif 3216	. . 3 |- (B (+) C) = ((B \ C) u. (C \ B))
10		df-un 3214	. . 3 |- ((B \ C) u. (C \ B)) = ( ∼ (B \ C) &ncap ∼ (C \ B))
11	9, 10	eqtri 2373	. 2 |- (B (+) C) = ( ∼ (B \ C) &ncap ∼ (C \ B))
12	5, 8, 11	3eqtr4g 2410	1 |- (A = B -> (A (+) C) = (B (+) C))

Syntax hints:   -> wi 4   = wceq 1642   &ncap cnin 3204   ∼ ccompl 3205   \ cdif 3206   u. cun 3207   (+) csymdif 3209

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 2478  df-v 2861  df-nin 3211  df-compl 3212  df-in 3213  df-un 3214  df-dif 3215  df-symdif 3216

This theorem is referenced by:  symdifeq12  3250  symdifeq1i  3251  symdifeq1d  3254