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Theorem tpeq3 3754
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 3677 . . 3 (𝐴 = 𝐵 → {𝐴} = {𝐵})
21uneq2d 3358 . 2 (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3 df-tp 3674 . 2 {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4 df-tp 3674 . 2 {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
52, 3, 43eqtr4g 2287 1 (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1395  cun 3195  {csn 3666  {cpr 3667  {ctp 3668
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211
This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2801  df-un 3201  df-sn 3672  df-tp 3674
This theorem is referenced by:  tpeq3d  3757  tppreq3  3769  fztpval  10275
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