Theorem tpeq1 3704

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 3695	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
2	1	uneq1d 3312	. 2 ⊢ (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3		df-tp 3626	. 2 ⊢ {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4		df-tp 3626	. 2 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2251	1 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})

Syntax hints:   → wi 4   = wceq 1364   ∪ cun 3151  {csn 3618  {cpr 3619  {ctp 3620

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 3157  df-sn 3624  df-pr 3625  df-tp 3626

This theorem is referenced by:  tpeq1d  3707