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Theorem tpeq3 3480
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 3409 . . 3 (𝐴 = 𝐵 → {𝐴} = {𝐵})
21uneq2d 3126 . 2 (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3 df-tp 3406 . 2 {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4 df-tp 3406 . 2 {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
52, 3, 43eqtr4g 2138 1 (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1284  cun 2971  {csn 3398  {cpr 3399  {ctp 3400
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063
This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-sn 3404  df-tp 3406
This theorem is referenced by:  tpeq3d  3483  tppreq3  3495  fztpval  9100
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