Theorem symdifeq1 4208

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 4074	. . 3 ⊢ (𝐴 = 𝐵 → (𝐴 ∖ 𝐶) = (𝐵 ∖ 𝐶))
2		difeq2 4075	. . 3 ⊢ (𝐴 = 𝐵 → (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))
3	1, 2	uneq12d 4123	. 2 ⊢ (𝐴 = 𝐵 → ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴)) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵)))
4		df-symdif 4206	. 2 ⊢ (𝐴 △ 𝐶) = ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴))
5		df-symdif 4206	. 2 ⊢ (𝐵 △ 𝐶) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵))
6	3, 4, 5	3eqtr4g 2825	1 ⊢ (𝐴 = 𝐵 → (𝐴 △ 𝐶) = (𝐵 △ 𝐶))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∖ cdif 3903   ∪ cun 3904   △ csymdif 4205

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-symdif 4206

This theorem is used by:  symdifeq2  4209