Theorem symdifeq1 4184

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 4055	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∖ 𝐶) = (𝐵 ∖ 𝐶))
2		difeq2 4056	. . 3 ⊢ (𝐴 = 𝐵 → (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))
3	1, 2	uneq12d 4103	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴)) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵)))
4		df-symdif 4182	. 2 ⊢ (𝐴 △ 𝐶) = ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴))
5		df-symdif 4182	. 2 ⊢ (𝐵 △ 𝐶) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵))
6	3, 4, 5	3eqtr4g 2805	1 ⊢ (𝐴 = 𝐵 → (𝐴 △ 𝐶) = (𝐵 △ 𝐶))

Syntax hints:   → wi 4   = wceq 1542   ∖ cdif 3889   ∪ cun 3890   △ csymdif 4181

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1545  df-ex 1787  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-rab 3075  df-v 3433  df-dif 3895  df-un 3897  df-symdif 4182

This theorem is referenced by:  symdifeq2  4185