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Theorem symdifeq1 4038
Description: Equality theorem for symmetric difference. (Contributed by Scott Fenton, 24-Apr-2012.)
Assertion
Ref Expression
symdifeq1 (𝐴 = 𝐵 → (𝐴𝐶) = (𝐵𝐶))

Proof of Theorem symdifeq1
StepHypRef Expression
1 difeq1 3914 . . 3 (𝐴 = 𝐵 → (𝐴𝐶) = (𝐵𝐶))
2 difeq2 3915 . . 3 (𝐴 = 𝐵 → (𝐶𝐴) = (𝐶𝐵))
31, 2uneq12d 3961 . 2 (𝐴 = 𝐵 → ((𝐴𝐶) ∪ (𝐶𝐴)) = ((𝐵𝐶) ∪ (𝐶𝐵)))
4 df-symdif 4036 . 2 (𝐴𝐶) = ((𝐴𝐶) ∪ (𝐶𝐴))
5 df-symdif 4036 . 2 (𝐵𝐶) = ((𝐵𝐶) ∪ (𝐶𝐵))
63, 4, 53eqtr4g 2861 1 (𝐴 = 𝐵 → (𝐴𝐶) = (𝐵𝐶))
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1637  cdif 3760  cun 3761  csymdif 4035
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1877  ax-4 1894  ax-5 2001  ax-6 2067  ax-7 2103  ax-9 2164  ax-10 2184  ax-11 2200  ax-12 2213  ax-13 2419  ax-ext 2781
This theorem depends on definitions:  df-bi 198  df-an 385  df-or 866  df-tru 1641  df-ex 1860  df-nf 1864  df-sb 2060  df-clab 2789  df-cleq 2795  df-clel 2798  df-nfc 2933  df-ral 3097  df-rab 3101  df-v 3389  df-dif 3766  df-un 3768  df-symdif 4036
This theorem is referenced by:  symdifeq2  4039
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