Theorem tpeq3 3707

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 3630	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴} = {𝐵})
2	1	uneq2d 3314	. 2 ⊢ (𝐴 = 𝐵 → ({𝐶, 𝐷} ∪ {𝐴}) = ({𝐶, 𝐷} ∪ {𝐵}))
3		df-tp 3627	. 2 ⊢ {𝐶, 𝐷, 𝐴} = ({𝐶, 𝐷} ∪ {𝐴})
4		df-tp 3627	. 2 ⊢ {𝐶, 𝐷, 𝐵} = ({𝐶, 𝐷} ∪ {𝐵})
5	2, 3, 4	3eqtr4g 2251	1 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐷, 𝐴} = {𝐶, 𝐷, 𝐵})

Syntax hints:   → wi 4   = wceq 1364   ∪ cun 3152  {csn 3619  {cpr 3620  {ctp 3621

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 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175

This theorem depends on definitions:  df-bi 117  df-tru 1367  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-v 2762  df-un 3158  df-sn 3625  df-tp 3627

This theorem is referenced by:  tpeq3d  3710  tppreq3  3722  fztpval  10152