Theorem tpeq1 3777

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

Step	Hyp	Ref	Expression
1		preq1 3768	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
2	1	uneq1d 3372	. 2 ⊢ (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3		df-tp 3697	. 2 ⊢ {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4		df-tp 3697	. 2 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2290	1 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})

Syntax hints:   → wi 4   = wceq 1398   ∪ cun 3209  {csn 3689  {cpr 3690  {ctp 3691

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 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-ext 2214

This theorem depends on definitions:  df-bi 117  df-tru 1401  df-nf 1510  df-sb 1812  df-clab 2219  df-cleq 2225  df-clel 2228  df-nfc 2373  df-v 2815  df-un 3215  df-sn 3695  df-pr 3696  df-tp 3697

This theorem is referenced by:  tpeq1d  3780