Theorem symdifeq1 4171

Description: Equality theorem for symmetric difference. (Contributed by Scott Fenton, 24-Apr-2012.)

Assertion

Ref	Expression
symdifeq1	⊢ (𝐴 = 𝐵 → (𝐴 △ 𝐶) = (𝐵 △ 𝐶))

Proof of Theorem symdifeq1

Step	Hyp	Ref	Expression
1		difeq1 4043	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∖ 𝐶) = (𝐵 ∖ 𝐶))
2		difeq2 4044	. . 3 ⊢ (𝐴 = 𝐵 → (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))
3	1, 2	uneq12d 4091	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴)) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵)))
4		df-symdif 4169	. 2 ⊢ (𝐴 △ 𝐶) = ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴))
5		df-symdif 4169	. 2 ⊢ (𝐵 △ 𝐶) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵))
6	3, 4, 5	3eqtr4g 2858	1 ⊢ (𝐴 = 𝐵 → (𝐴 △ 𝐶) = (𝐵 △ 𝐶))

Syntax hints:   → wi 4   = wceq 1538   ∖ cdif 3878   ∪ cun 3879   △ csymdif 4168

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2113  ax-9 2121  ax-ext 2770

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 845  df-ex 1782  df-sb 2070  df-clab 2777  df-cleq 2791  df-clel 2870  df-rab 3115  df-v 3443  df-dif 3884  df-un 3886  df-symdif 4169

This theorem is referenced by:  symdifeq2  4172