Theorem opthpr 3774

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 3772	. 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 2348	. . . . . 6 ⊢ (𝐴 ≠ 𝐷 ↔ ¬ 𝐴 = 𝐷)
8		pm2.21 617	. . . . . 6 ⊢ (¬ 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
9	7, 8	sylbi 121	. . . . 5 ⊢ (𝐴 ≠ 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
10	9	impd 254	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
11	6, 10	jaod 717	. . 3 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
12		orc 712	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	11, 12	impbid1 142	. 2 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
14	5, 13	bitrid 192	1 ⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   ∨ wo 708   = wceq 1353   ∈ wcel 2148   ≠ wne 2347  Vcvv 2739  {cpr 3595

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ne 2348  df-v 2741  df-un 3135  df-sn 3600  df-pr 3601

This theorem is referenced by: (None)