Theorem tpeq1 3724

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 3715	. . 3 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
2	1	uneq1d 3330	. 2 ⊢ (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3		df-tp 3646	. 2 ⊢ {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4		df-tp 3646	. 2 ⊢ {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
5	2, 3, 4	3eqtr4g 2264	1 ⊢ (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})

Syntax hints:   → wi 4   = wceq 1373   ∪ cun 3168  {csn 3638  {cpr 3639  {ctp 3640

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-un 3174  df-sn 3644  df-pr 3645  df-tp 3646

This theorem is referenced by:  tpeq1d  3727