ILE Home Intuitionistic Logic Explorer < Previous   Next >
Nearby theorems
Mirrors  >  Home  >  ILE Home  >  Th. List  >  preqsn GIF version

Theorem preqsn 3762
Description: Equivalence for a pair equal to a singleton. (Contributed by NM, 3-Jun-2008.)
Hypotheses
Ref Expression
preqsn.1 𝐴 ∈ V
preqsn.2 𝐵 ∈ V
preqsn.3 𝐶 ∈ V
Assertion
Ref Expression
preqsn ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐵𝐵 = 𝐶))

Proof of Theorem preqsn
StepHypRef Expression
1 dfsn2 3597 . . 3 {𝐶} = {𝐶, 𝐶}
21eqeq2i 2181 . 2 ({𝐴, 𝐵} = {𝐶} ↔ {𝐴, 𝐵} = {𝐶, 𝐶})
3 preqsn.1 . . . 4 𝐴 ∈ V
4 preqsn.2 . . . 4 𝐵 ∈ V
5 preqsn.3 . . . 4 𝐶 ∈ V
63, 4, 5, 5preq12b 3757 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ ((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)))
7 oridm 752 . . . 4 (((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐶))
8 eqtr3 2190 . . . . . 6 ((𝐴 = 𝐶𝐵 = 𝐶) → 𝐴 = 𝐵)
9 simpr 109 . . . . . 6 ((𝐴 = 𝐶𝐵 = 𝐶) → 𝐵 = 𝐶)
108, 9jca 304 . . . . 5 ((𝐴 = 𝐶𝐵 = 𝐶) → (𝐴 = 𝐵𝐵 = 𝐶))
11 eqtr 2188 . . . . . 6 ((𝐴 = 𝐵𝐵 = 𝐶) → 𝐴 = 𝐶)
12 simpr 109 . . . . . 6 ((𝐴 = 𝐵𝐵 = 𝐶) → 𝐵 = 𝐶)
1311, 12jca 304 . . . . 5 ((𝐴 = 𝐵𝐵 = 𝐶) → (𝐴 = 𝐶𝐵 = 𝐶))
1410, 13impbii 125 . . . 4 ((𝐴 = 𝐶𝐵 = 𝐶) ↔ (𝐴 = 𝐵𝐵 = 𝐶))
157, 14bitri 183 . . 3 (((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)) ↔ (𝐴 = 𝐵𝐵 = 𝐶))
166, 15bitri 183 . 2 ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ (𝐴 = 𝐵𝐵 = 𝐶))
172, 16bitri 183 1 ({𝐴, 𝐵} = {𝐶} ↔ (𝐴 = 𝐵𝐵 = 𝐶))
Colors of variables: wff set class
Syntax hints:  wa 103  wb 104  wo 703   = wceq 1348  wcel 2141  Vcvv 2730  {csn 3583  {cpr 3584
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 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152
This theorem depends on definitions:  df-bi 116  df-tru 1351  df-nf 1454  df-sb 1756  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732  df-un 3125  df-sn 3589  df-pr 3590
This theorem is referenced by:  opeqsn  4237  relop  4761
  Copyright terms: Public domain W3C validator