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Theorem tpeq3 3449
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
StepHypRef Expression
1 sneq 3378 . . 3 (A = B → {A} = {B})
21uneq2d 3091 . 2 (A = B → ({𝐶, 𝐷} ∪ {A}) = ({𝐶, 𝐷} ∪ {B}))
3 df-tp 3375 . 2 {𝐶, 𝐷, A} = ({𝐶, 𝐷} ∪ {A})
4 df-tp 3375 . 2 {𝐶, 𝐷, B} = ({𝐶, 𝐷} ∪ {B})
52, 3, 43eqtr4g 2094 1 (A = B → {𝐶, 𝐷, A} = {𝐶, 𝐷, B})
Colors of variables: wff set class
Syntax hints:  wi 4   = wceq 1242  cun 2909  {csn 3367  {cpr 3368  {ctp 3369
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 629  ax-5 1333  ax-7 1334  ax-gen 1335  ax-ie1 1379  ax-ie2 1380  ax-8 1392  ax-10 1393  ax-11 1394  ax-i12 1395  ax-bnd 1396  ax-4 1397  ax-17 1416  ax-i9 1420  ax-ial 1424  ax-i5r 1425  ax-ext 2019
This theorem depends on definitions:  df-bi 110  df-tru 1245  df-nf 1347  df-sb 1643  df-clab 2024  df-cleq 2030  df-clel 2033  df-nfc 2164  df-v 2553  df-un 2916  df-sn 3373  df-tp 3375
This theorem is referenced by:  tpeq3d  3452  tppreq3  3464  fztpval  8715
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