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Theorem tpeq3 3810
 Description: Equality theorem for unordered triples. (Contributed by NM, 13-Sep-2011.)
Assertion
Ref Expression
tpeq3 (A = B → {C, D, A} = {C, D, B})

Proof of Theorem tpeq3
StepHypRef Expression
1 sneq 3744 . . 3 (A = B → {A} = {B})
21uneq2d 3418 . 2 (A = B → ({C, D} ∪ {A}) = ({C, D} ∪ {B}))
3 df-tp 3743 . 2 {C, D, A} = ({C, D} ∪ {A})
4 df-tp 3743 . 2 {C, D, B} = ({C, D} ∪ {B})
52, 3, 43eqtr4g 2410 1 (A = B → {C, D, A} = {C, D, B})
 Colors of variables: wff setvar class Syntax hints:   → wi 4   = wceq 1642   ∪ cun 3207  {csn 3737  {cpr 3738  {ctp 3739 This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-3 7  ax-mp 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334 This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2478  df-v 2861  df-nin 3211  df-compl 3212  df-un 3214  df-sn 3741  df-tp 3743 This theorem is referenced by:  tpeq3d  3813
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