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Theorem tpeq1 4675
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 4666 . . 3 (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
21uneq1d 4092 . 2 (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3 df-tp 4563 . 2 {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4 df-tp 4563 . 2 {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
52, 3, 43eqtr4g 2804 1 (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})
Colors of variables: wff setvar class
Syntax hints:  wi 4   = wceq 1539  cun 3881  {csn 4558  {cpr 4560  {ctp 4562
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709
This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-v 3424  df-un 3888  df-sn 4559  df-pr 4561  df-tp 4563
This theorem is referenced by:  tpeq1d  4678  hashtpg  14127  erngset  38741  erngset-rN  38749  dvh4dimN  39388  lmod1  45721
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