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) ⩃ ∼ (C ∖ A)) = ( ∼ (B ∖ C) ⩃ ∼ (C ∖ B)))
6		df-symdif 3216	. . 3 ⊢ (A ⊕ C) = ((A ∖ C) ∪ (C ∖ A))
7		df-un 3214	. . 3 ⊢ ((A ∖ C) ∪ (C ∖ A)) = ( ∼ (A ∖ C) ⩃ ∼ (C ∖ A))
8	6, 7	eqtri 2373	. 2 ⊢ (A ⊕ C) = ( ∼ (A ∖ C) ⩃ ∼ (C ∖ A))
9		df-symdif 3216	. . 3 ⊢ (B ⊕ C) = ((B ∖ C) ∪ (C ∖ B))
10		df-un 3214	. . 3 ⊢ ((B ∖ C) ∪ (C ∖ B)) = ( ∼ (B ∖ C) ⩃ ∼ (C ∖ B))
11	9, 10	eqtri 2373	. 2 ⊢ (B ⊕ C) = ( ∼ (B ∖ C) ⩃ ∼ (C ∖ B))
12	5, 8, 11	3eqtr4g 2410	1 ⊢ (A = B → (A ⊕ C) = (B ⊕ C))

Syntax hints:   → wi 4   = wceq 1642   ⩃ cnin 3204   ∼ ccompl 3205   ∖ cdif 3206   ∪ 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