Theorem preqr1 4125

Description: Reverse equality lemma for unordered pairs. If two unordered pairs have the same second element, the first elements are equal. (Contributed by NM, 18-Oct-1995.)

Hypotheses

Ref	Expression
preqr1.1	⊢ A ∈ V
preqr1.2	⊢ B ∈ V

Assertion

Ref	Expression
preqr1	⊢ ({A, C} = {B, C} → A = B)

Proof of Theorem preqr1

Step	Hyp	Ref	Expression
1		preqr1.1	. . . . 5 ⊢ A ∈ V
2	1	prid1 3828	. . . 4 ⊢ A ∈ {A, C}
3		eleq2 2414	. . . 4 ⊢ ({A, C} = {B, C} → (A ∈ {A, C} ↔ A ∈ {B, C}))
4	2, 3	mpbii 202	. . 3 ⊢ ({A, C} = {B, C} → A ∈ {B, C})
5	1	elpr 3752	. . 3 ⊢ (A ∈ {B, C} ↔ (A = B ∨ A = C))
6	4, 5	sylib 188	. 2 ⊢ ({A, C} = {B, C} → (A = B ∨ A = C))
7		preqr1.2	. . . . 5 ⊢ B ∈ V
8	7	prid1 3828	. . . 4 ⊢ B ∈ {B, C}
9		eleq2 2414	. . . 4 ⊢ ({A, C} = {B, C} → (B ∈ {A, C} ↔ B ∈ {B, C}))
10	8, 9	mpbiri 224	. . 3 ⊢ ({A, C} = {B, C} → B ∈ {A, C})
11	7	elpr 3752	. . 3 ⊢ (B ∈ {A, C} ↔ (B = A ∨ B = C))
12	10, 11	sylib 188	. 2 ⊢ ({A, C} = {B, C} → (B = A ∨ B = C))
13		eqcom 2355	. 2 ⊢ (A = B ↔ B = A)
14		eqeq2 2362	. 2 ⊢ (A = C → (B = A ↔ B = C))
15	6, 12, 13, 14	oplem1 930	1 ⊢ ({A, C} = {B, C} → A = B)

Syntax hints:   → wi 4   ∨ wo 357   = wceq 1642   ∈ wcel 1710  Vcvv 2860  {cpr 3739

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1546  ax-5 1557  ax-17 1616  ax-9 1654  ax-8 1675  ax-6 1729  ax-7 1734  ax-11 1746  ax-12 1925  ax-ext 2334

This theorem depends on definitions:  df-bi 177  df-or 359  df-an 360  df-nan 1288  df-tru 1319  df-ex 1542  df-nf 1545  df-sb 1649  df-clab 2340  df-cleq 2346  df-clel 2349  df-nfc 2479  df-v 2862  df-nin 3212  df-compl 3213  df-un 3215  df-sn 3742  df-pr 3743

This theorem is referenced by:  preqr2  4126