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Theorem symdifeq1 4203
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
StepHypRef Expression
1 difeq1 4059 . . 3 (𝐴 = 𝐵 → (𝐴 ∖ 𝐶) = (𝐵 ∖ 𝐶))
2 difeq2 4060 . . 3 (𝐴 = 𝐵 → (𝐶 ∖ 𝐴) = (𝐶 ∖ 𝐵))
31, 2uneq12d 4171 . 2 (𝐴 = 𝐵 → ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴)) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵)))
4 df-symdif 4201 . 2 (𝐴 △ 𝐶) = ((𝐴 ∖ 𝐶) ∪ (𝐶 ∖ 𝐴))
5 df-symdif 4201 . 2 (𝐵 △ 𝐶) = ((𝐵 ∖ 𝐶) ∪ (𝐶 ∖ 𝐵))
63, 4, 53eqtr4g 2823 1 (𝐴 = 𝐵 → (𝐴 △ 𝐶) = (𝐵 △ 𝐶))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   = wceq 1570   ∖ cdif 4048   ∪ cun 4152   △ csymdif 4200
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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-rab 3417  df-v 3456  df-dif 4050  df-un 4154  df-symdif 4201
This theorem is used by:  symdifeq2  4204
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