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Theorem tpeq3 3795
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 3716 . . 3 (𝐴 = 𝐵 → {𝐴} = {𝐵})
21uneq2d 3383 . 2 (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3 df-tp 3713 . 2 {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4 df-tp 3713 . 2 {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
52, 3, 43eqtr4g 2296 1 (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1402  cun 3218  {csn 3705  {cpr 3706  {ctp 3707
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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220
This theorem depends on definitions:  df-bi 117  df-tru 1405  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-v 2823  df-un 3224  df-sn 3711  df-tp 3713
This theorem is referenced by:  tpeq3d  3798  tppreq3  3810  fztpval  10468
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