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Theorem tpeq1 3609
 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
StepHypRef Expression
1 preq1 3600 . . 3 (𝐴 = 𝐵 → {𝐴, 𝐶} = {𝐵, 𝐶})
21uneq1d 3229 . 2 (𝐴 = 𝐵 → ({𝐴, 𝐶} ∪ {𝐷}) = ({𝐵, 𝐶} ∪ {𝐷}))
3 df-tp 3535 . 2 {𝐴, 𝐶, 𝐷} = ({𝐴, 𝐶} ∪ {𝐷})
4 df-tp 3535 . 2 {𝐵, 𝐶, 𝐷} = ({𝐵, 𝐶} ∪ {𝐷})
52, 3, 43eqtr4g 2197 1 (𝐴 = 𝐵 → {𝐴, 𝐶, 𝐷} = {𝐵, 𝐶, 𝐷})
 Colors of variables: wff set class Syntax hints:   → wi 4   = wceq 1331   ∪ cun 3069  {csn 3527  {cpr 3528  {ctp 3529 This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121 This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-un 3075  df-sn 3533  df-pr 3534  df-tp 3535 This theorem is referenced by:  tpeq1d  3612
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