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Theorem opeqpr 4015
Description: Equivalence for an ordered pair equal to an unordered pair. (Contributed by NM, 3-Jun-2008.)
Hypotheses
Ref Expression
opeqpr.1 𝐴 ∈ V
opeqpr.2 𝐵 ∈ V
opeqpr.3 𝐶 ∈ V
opeqpr.4 𝐷 ∈ V
Assertion
Ref Expression
opeqpr (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))

Proof of Theorem opeqpr
StepHypRef Expression
1 eqcom 2056 . 2 (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ {𝐶, 𝐷} = ⟨𝐴, 𝐵⟩)
2 opeqpr.1 . . . 4 𝐴 ∈ V
3 opeqpr.2 . . . 4 𝐵 ∈ V
42, 3dfop 3573 . . 3 𝐴, 𝐵⟩ = {{𝐴}, {𝐴, 𝐵}}
54eqeq2i 2064 . 2 ({𝐶, 𝐷} = ⟨𝐴, 𝐵⟩ ↔ {𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}})
6 opeqpr.3 . . 3 𝐶 ∈ V
7 opeqpr.4 . . 3 𝐷 ∈ V
8 snexgOLD 3960 . . . 4 (𝐴 ∈ V → {𝐴} ∈ V)
92, 8ax-mp 7 . . 3 {𝐴} ∈ V
10 prexgOLD 3971 . . . 4 ((𝐴 ∈ V ∧ 𝐵 ∈ V) → {𝐴, 𝐵} ∈ V)
112, 3, 10mp2an 410 . . 3 {𝐴, 𝐵} ∈ V
126, 7, 9, 11preq12b 3566 . 2 ({𝐶, 𝐷} = {{𝐴}, {𝐴, 𝐵}} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
131, 5, 123bitri 199 1 (⟨𝐴, 𝐵⟩ = {𝐶, 𝐷} ↔ ((𝐶 = {𝐴} ∧ 𝐷 = {𝐴, 𝐵}) ∨ (𝐶 = {𝐴, 𝐵} ∧ 𝐷 = {𝐴})))
Colors of variables: wff set class
Syntax hints:  wa 101  wb 102  wo 637   = wceq 1257  wcel 1407  Vcvv 2572  {csn 3400  {cpr 3401  cop 3403
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-io 638  ax-5 1350  ax-7 1351  ax-gen 1352  ax-ie1 1396  ax-ie2 1397  ax-8 1409  ax-10 1410  ax-11 1411  ax-i12 1412  ax-bndl 1413  ax-4 1414  ax-14 1419  ax-17 1433  ax-i9 1437  ax-ial 1441  ax-i5r 1442  ax-ext 2036  ax-sep 3900  ax-pow 3952  ax-pr 3969
This theorem depends on definitions:  df-bi 114  df-3an 896  df-tru 1260  df-nf 1364  df-sb 1660  df-clab 2041  df-cleq 2047  df-clel 2050  df-nfc 2181  df-v 2574  df-un 2947  df-in 2949  df-ss 2956  df-pw 3386  df-sn 3406  df-pr 3407  df-op 3409
This theorem is referenced by:  relop  4511
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