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Theorem tpeq1 3809
Description: Equality theorem for unordered triples. (Contributed by NM, 13-Sep-2011.)
Assertion
Ref Expression
tpeq1 (A = B → {A, C, D} = {B, C, D})

Proof of Theorem tpeq1
StepHypRef Expression
1 preq1 3800 . . 3 (A = B → {A, C} = {B, C})
21uneq1d 3418 . 2 (A = B → ({A, C} ∪ {D}) = ({B, C} ∪ {D}))
3 df-tp 3744 . 2 {A, C, D} = ({A, C} ∪ {D})
4 df-tp 3744 . 2 {B, C, D} = ({B, C} ∪ {D})
52, 3, 43eqtr4g 2410 1 (A = B → {A, C, D} = {B, C, D})
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1642  cun 3208  {csn 3738  {cpr 3739  {ctp 3740
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334
This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215  df-sn 3742  df-pr 3743  df-tp 3744
This theorem is referenced by:  tpeq1d  3812
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