Theorem tpeq3 3611

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

Assertion

Ref	Expression
tpeq3	⊢ (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})

Proof of Theorem tpeq3

Step	Hyp	Ref	Expression
1		sneq 3538	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴} = {𝐵})
2	1	uneq2d 3230	. 2 ⊢ (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3		df-tp 3535	. 2 ⊢ {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4		df-tp 3535	. 2 ⊢ {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
5	2, 3, 4	3eqtr4g 2197	1 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})

Syntax hints:   → wi 4   = wceq 1331   ∪ cun 3069  {csn 3527  {cpr 3528  {ctp 3529

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-un 3075  df-sn 3533  df-tp 3535

This theorem is referenced by:  tpeq3d  3614  tppreq3  3626  fztpval  9863