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Theorem preqsn 3755
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 3590 . . 3 {𝐶} = {𝐶, 𝐶}
21eqeq2i 2176 . 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 3750 . . 3 ({𝐴, 𝐵} = {𝐶, 𝐶} ↔ ((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)))
7 oridm 747 . . . 4 (((𝐴 = 𝐶𝐵 = 𝐶) ∨ (𝐴 = 𝐶𝐵 = 𝐶)) ↔ (𝐴 = 𝐶𝐵 = 𝐶))
8 eqtr3 2185 . . . . . 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 2183 . . . . . 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 698   = wceq 1343  wcel 2136  Vcvv 2726  {csn 3576  {cpr 3577
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 699  ax-5 1435  ax-7 1436  ax-gen 1437  ax-ie1 1481  ax-ie2 1482  ax-8 1492  ax-10 1493  ax-11 1494  ax-i12 1495  ax-bndl 1497  ax-4 1498  ax-17 1514  ax-i9 1518  ax-ial 1522  ax-i5r 1523  ax-ext 2147
This theorem depends on definitions:  df-bi 116  df-tru 1346  df-nf 1449  df-sb 1751  df-clab 2152  df-cleq 2158  df-clel 2161  df-nfc 2297  df-v 2728  df-un 3120  df-sn 3582  df-pr 3583
This theorem is referenced by:  opeqsn  4230  relop  4754
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