Theorem tpeq2 3479

Description: Equality theorem for unordered triples. (Contributed by NM, 13-Sep-2011.)

Assertion

Ref	Expression
tpeq2	⊢ (𝐴 = 𝐵 → {𝐶, 𝐴, 𝐷} = {𝐶, 𝐵, 𝐷})

Proof of Theorem tpeq2

Step	Hyp	Ref	Expression
1		preq2 3470	. . 3 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐴} = {𝐶, 𝐵})
2	1	uneq1d 3125	. 2 ⊢ (𝐴 = 𝐵 → ({𝐶, 𝐴} ∪ {𝐷}) = ({𝐶, 𝐵} ∪ {𝐷}))
3		df-tp 3406	. 2 ⊢ {𝐶, 𝐴, 𝐷} = ({𝐶, 𝐴} ∪ {𝐷})
4		df-tp 3406	. 2 ⊢ {𝐶, 𝐵, 𝐷} = ({𝐶, 𝐵} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2138	1 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐴, 𝐷} = {𝐶, 𝐵, 𝐷})

Syntax hints:   → wi 4   = wceq 1284   ∪ cun 2971  {csn 3398  {cpr 3399  {ctp 3400

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-v 2603  df-un 2977  df-sn 3404  df-pr 3405  df-tp 3406

This theorem is referenced by:  tpeq2d  3482  fztpval  9100