Theorem opthpr 4811

Description: An unordered pair has the ordered pair property (compare opth 5445) 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 4810	. 2 ⊢ ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
6		idd 25	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
7		df-ne 2957	. . . . . 6 ⊢ (𝐴 ≠ 𝐷 ↔ ¬ 𝐴 = 𝐷)
8		pm2.21 124	. . . . . 6 ⊢ (¬ 𝐴 = 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
9	7, 8	sylbi 220	. . . . 5 ⊢ (𝐴 ≠ 𝐷 → (𝐴 = 𝐷 → (𝐵 = 𝐶 → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))))
10	9	impd 416	. . . 4 ⊢ (𝐴 ≠ 𝐷 → ((𝐴 = 𝐷 ∧ 𝐵 = 𝐶) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
11	6, 10	jaod 873	. . 3 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
12		orc 881	. . 3 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)))
13	11, 12	impbid1 228	. 2 ⊢ (𝐴 ≠ 𝐷 → (((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) ∨ (𝐴 = 𝐷 ∧ 𝐵 = 𝐶)) ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))
14	5, 13	bitrid 286	1 ⊢ (𝐴 ≠ 𝐷 → ({𝐴, 𝐵} = {𝐶, 𝐷} ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   ∨ wo 861   = wceq 1570   ∈ wcel 2145   ≠ wne 2956  Vcvv 3451  {cpr 4586

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-ext 2733

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-tru 1573  df-ex 1813  df-sb 2100  df-clab 2740  df-cleq 2753  df-clel 2836  df-ne 2957  df-v 3453  df-un 3904  df-sn 4585  df-pr 4587

This theorem is used by:  brdom7disj  10610  brdom6disj  10611