Theorem opthpr 4809

Description: An unordered pair has the ordered pair property (compare opth 5444) under certain conditions. (Contributed by NM, 27-Mar-2007.)

Hypotheses

Ref	Expression
preqr1.a	⊢ 𝐴 ∈ V
preqr1.b	⊢ 𝐵 ∈ V
preq12b.c	⊢ 𝐶 ∈ V
preq12b.d	⊢ 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		preqr1.a	. . 3 ⊢ 𝐴 ∈ V
2		preqr1.b	. . 3 ⊢ 𝐵 ∈ V
3		preq12b.c	. . 3 ⊢ 𝐶 ∈ V
4		preq12b.d	. . 3 ⊢ 𝐷 ∈ V
5	1, 2, 3, 4	preq12b 4808	. 2 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6		idd 24	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
7		df-ne 2958	. . . . . 6 ⊢ (𝐴 ≠ 𝐷 ↔ ¬ 𝐴 = 𝐷)
8		pm2.21 123	. . . . . 6 ⊢ (¬ 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
9	7, 8	sylbi 219	. . . . 5 ⊢ (𝐴 ≠ 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
10	9	impd 414	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
11	6, 10	jaod 870	. . 3 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
12		orc 878	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	11, 12	impbid1 227	. 2 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
14	5, 13	bitrid 285	1 ⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399   ∨ wo 858   = wceq 1560   ∈ wcel 2142   ≠ wne 2957  Vcvv 3454  {cpr 4584

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1815  ax-4 1829  ax-5 1930  ax-6 1987  ax-7 2028  ax-8 2144  ax-9 2152  ax-ext 2734

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1563  df-ex 1800  df-sb 2091  df-clab 2741  df-cleq 2754  df-clel 2837  df-ne 2958  df-v 3456  df-un 3909  df-sn 4583  df-pr 4585

This theorem is referenced by:  brdom7disj  10488  brdom6disj  10489