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) ⩃ ∼ (A ∖ C)) = ( ∼ (C ∖ B) ⩃ ∼ (B ∖ C)))
6		df-symdif 3217	. . 3 ⊢ (C ⊕ A) = ((C ∖ A) ∪ (A ∖ C))
7		df-un 3215	. . 3 ⊢ ((C ∖ A) ∪ (A ∖ C)) = ( ∼ (C ∖ A) ⩃ ∼ (A ∖ C))
8	6, 7	eqtri 2373	. 2 ⊢ (C ⊕ A) = ( ∼ (C ∖ A) ⩃ ∼ (A ∖ C))
9		df-symdif 3217	. . 3 ⊢ (C ⊕ B) = ((C ∖ B) ∪ (B ∖ C))
10		df-un 3215	. . 3 ⊢ ((C ∖ B) ∪ (B ∖ C)) = ( ∼ (C ∖ B) ⩃ ∼ (B ∖ C))
11	9, 10	eqtri 2373	. 2 ⊢ (C ⊕ B) = ( ∼ (C ∖ B) ⩃ ∼ (B ∖ C))
12	5, 8, 11	3eqtr4g 2410	1 ⊢ (A = B → (C ⊕ A) = (C ⊕ B))

Syntax hints:   → wi 4   = wceq 1642   ⩃ cnin 3205   ∼ ccompl 3206   ∖ cdif 3207   ∪ 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