Theorem opthpr 4779

Description: An unordered pair has the ordered pair property (compare opth 5385) 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 4778	. 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 2943	. . . . . 6 ⊢ (𝐴 ≠ 𝐷 ↔ ¬ 𝐴 = 𝐷)
8		pm2.21 123	. . . . . 6 ⊢ (¬ 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
9	7, 8	sylbi 216	. . . . 5 ⊢ (𝐴 ≠ 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
10	9	impd 410	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
11	6, 10	jaod 855	. . 3 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
12		orc 863	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	11, 12	impbid1 224	. 2 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
14	5, 13	syl5bb 282	1 ⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 395   ∨ wo 843   = wceq 1539   ∈ wcel 2108   ≠ wne 2942  Vcvv 3422  {cpr 4560

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1799  ax-4 1813  ax-5 1914  ax-6 1972  ax-7 2012  ax-8 2110  ax-9 2118  ax-ext 2709

This theorem depends on definitions:  df-bi 206  df-an 396  df-or 844  df-tru 1542  df-ex 1784  df-sb 2069  df-clab 2716  df-cleq 2730  df-clel 2817  df-ne 2943  df-v 3424  df-un 3888  df-sn 4559  df-pr 4561

This theorem is referenced by:  brdom7disj  10218  brdom6disj  10219