Theorem tpeq2 3758

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 3749	. . 3 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐴} = {𝐶, 𝐵})
2	1	uneq1d 3360	. 2 ⊢ (𝐴 = 𝐵 → ({𝐶, 𝐴} ∪ {𝐷}) = ({𝐶, 𝐵} ∪ {𝐷}))
3		df-tp 3677	. 2 ⊢ {𝐶, 𝐴, 𝐷} = ({𝐶, 𝐴} ∪ {𝐷})
4		df-tp 3677	. 2 ⊢ {𝐶, 𝐵, 𝐷} = ({𝐶, 𝐵} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2289	1 ⊢ (𝐴 = 𝐵 → {𝐶, 𝐴, 𝐷} = {𝐶, 𝐵, 𝐷})

Syntax hints:   → wi 4   = wceq 1397   ∪ cun 3198  {csn 3669  {cpr 3670  {ctp 3671

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 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-un 3204  df-sn 3675  df-pr 3676  df-tp 3677

This theorem is referenced by:  tpeq2d  3761  fztpval  10317