Theorem tpeq1 3718

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 3709	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
2	1	uneq1d 3325	. 2 ⊢ (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3		df-tp 3640	. 2 ⊢ {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4		df-tp 3640	. 2 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2262	1 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})

Syntax hints:   → wi 4   = wceq 1372   ∪ cun 3163  {csn 3632  {cpr 3633  {ctp 3634

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 710  ax-5 1469  ax-7 1470  ax-gen 1471  ax-ie1 1515  ax-ie2 1516  ax-8 1526  ax-10 1527  ax-11 1528  ax-i12 1529  ax-bndl 1531  ax-4 1532  ax-17 1548  ax-i9 1552  ax-ial 1556  ax-i5r 1557  ax-ext 2186

This theorem depends on definitions:  df-bi 117  df-tru 1375  df-nf 1483  df-sb 1785  df-clab 2191  df-cleq 2197  df-clel 2200  df-nfc 2336  df-v 2773  df-un 3169  df-sn 3638  df-pr 3639  df-tp 3640

This theorem is referenced by:  tpeq1d  3721