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Theorem tpeq3 3525
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 3452 . . 3 (𝐴 = 𝐵 → {𝐴} = {𝐵})
21uneq2d 3152 . 2 (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3 df-tp 3449 . 2 {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4 df-tp 3449 . 2 {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
52, 3, 43eqtr4g 2145 1 (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1289  cun 2995  {csn 3441  {cpr 3442  {ctp 3443
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 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-bndl 1444  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-tru 1292  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-nfc 2217  df-v 2621  df-un 3001  df-sn 3447  df-tp 3449
This theorem is referenced by:  tpeq3d  3528  tppreq3  3540  fztpval  9464
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