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Theorem tpeq3 3671
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 3594 . . 3 (𝐴 = 𝐵 → {𝐴} = {𝐵})
21uneq2d 3281 . 2 (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3 df-tp 3591 . 2 {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4 df-tp 3591 . 2 {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
52, 3, 43eqtr4g 2228 1 (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1348  cun 3119  {csn 3583  {cpr 3584  {ctp 3585
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125  df-sn 3589  df-tp 3591
This theorem is referenced by:  tpeq3d  3674  tppreq3  3686  fztpval  10039
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