Theorem opthpr 3564

Description: A way to represent ordered pairs using unordered pairs with distinct members. (Contributed by NM, 27-Mar-2007.)

Hypotheses

Ref	Expression
preq12b.1	⊢ 𝐴 ∈ V
preq12b.2	⊢ 𝐵 ∈ V
preq12b.3	⊢ 𝐶 ∈ V
preq12b.4	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
opthpr	⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Proof of Theorem opthpr

Step	Hyp	Ref	Expression
1		preq12b.1	. . 3 ⊢ 𝐴 ∈ V
2		preq12b.2	. . 3 ⊢ 𝐵 ∈ V
3		preq12b.3	. . 3 ⊢ 𝐶 ∈ V
4		preq12b.4	. . 3 ⊢ 𝐷 ∈ V
5	1, 2, 3, 4	preq12b 3562	. 2 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6		idd 21	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
7		df-ne 2246	. . . . . 6 ⊢ (𝐴 ≠ 𝐷 ↔ ¬ 𝐴 = 𝐷)
8		pm2.21 579	. . . . . 6 ⊢ (¬ 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
9	7, 8	sylbi 119	. . . . 5 ⊢ (𝐴 ≠ 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
10	9	impd 251	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
11	6, 10	jaod 669	. . 3 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
12		orc 665	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	11, 12	impbid1 140	. 2 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
14	5, 13	syl5bb 190	1 ⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 102   ↔ wb 103   ∨ wo 661   = wceq 1284   ∈ wcel 1433   ≠ wne 2245  Vcvv 2601  {cpr 3399

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-in2 577  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063

This theorem depends on definitions:  df-bi 115  df-tru 1287  df-nf 1390  df-sb 1686  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ne 2246  df-v 2603  df-un 2977  df-sn 3404  df-pr 3405

This theorem is referenced by: (None)