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Theorem qdass 4062
Description: Two ways to write an unordered quadruple. (Contributed by Mario Carneiro, 5-Jan-2016.)
Assertion
Ref Expression
qdass  |-  ( { A ,  B }  u.  { C ,  D } )  =  ( { A ,  B ,  C }  u.  { D } )

Proof of Theorem qdass
StepHypRef Expression
1 unass 3582 . 2  |-  ( ( { A ,  B }  u.  { C } )  u.  { D } )  =  ( { A ,  B }  u.  ( { C }  u.  { D } ) )
2 df-tp 3964 . . 3  |-  { A ,  B ,  C }  =  ( { A ,  B }  u.  { C } )
32uneq1i 3575 . 2  |-  ( { A ,  B ,  C }  u.  { D } )  =  ( ( { A ,  B }  u.  { C } )  u.  { D } )
4 df-pr 3962 . . 3  |-  { C ,  D }  =  ( { C }  u.  { D } )
54uneq2i 3576 . 2  |-  ( { A ,  B }  u.  { C ,  D } )  =  ( { A ,  B }  u.  ( { C }  u.  { D } ) )
61, 3, 53eqtr4ri 2504 1  |-  ( { A ,  B }  u.  { C ,  D } )  =  ( { A ,  B ,  C }  u.  { D } )
Colors of variables: wff setvar class
Syntax hints:    = wceq 1452    u. cun 3388   {csn 3959   {cpr 3961   {ctp 3963
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1677  ax-4 1690  ax-5 1766  ax-6 1813  ax-7 1859  ax-10 1932  ax-11 1937  ax-12 1950  ax-13 2104  ax-ext 2451
This theorem depends on definitions:  df-bi 190  df-or 377  df-an 378  df-tru 1455  df-ex 1672  df-nf 1676  df-sb 1806  df-clab 2458  df-cleq 2464  df-clel 2467  df-nfc 2601  df-v 3033  df-un 3395  df-pr 3962  df-tp 3964
This theorem is referenced by:  ex-pw  25958
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