Theorem tpeq1 3755

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 3746	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
2	1	uneq1d 3358	. 2 ⊢ (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3		df-tp 3675	. 2 ⊢ {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4		df-tp 3675	. 2 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2287	1 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})

Syntax hints:   → wi 4   = wceq 1395   ∪ cun 3196  {csn 3667  {cpr 3668  {ctp 3669

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 714  ax-5 1493  ax-7 1494  ax-gen 1495  ax-ie1 1539  ax-ie2 1540  ax-8 1550  ax-10 1551  ax-11 1552  ax-i12 1553  ax-bndl 1555  ax-4 1556  ax-17 1572  ax-i9 1576  ax-ial 1580  ax-i5r 1581  ax-ext 2211

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-nf 1507  df-sb 1809  df-clab 2216  df-cleq 2222  df-clel 2225  df-nfc 2361  df-v 2802  df-un 3202  df-sn 3673  df-pr 3674  df-tp 3675

This theorem is referenced by:  tpeq1d  3758