Theorem preqr1 3851

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 3777	. . . 4 ⊢ 𝐴 ∈ {𝐴, 𝐶}
3		eleq2 2295	. . . 4 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 ∈ {𝐴, 𝐶} ↔ 𝐴 ∈ {𝐵, 𝐶}))
4	2, 3	mpbii 148	. . 3 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 ∈ {𝐵, 𝐶})
5	1	elpr 3690	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
6	4, 5	sylib 122	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
7		preqr1.2	. . . . 5 ⊢ 𝐵 ∈ V
8	7	prid1 3777	. . . 4 ⊢ 𝐵 ∈ {𝐵, 𝐶}
9		eleq2 2295	. . . 4 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 ∈ {𝐴, 𝐶} ↔ 𝐵 ∈ {𝐵, 𝐶}))
10	8, 9	mpbiri 168	. . 3 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐵 ∈ {𝐴, 𝐶})
11	7	elpr 3690	. . 3 ⊢ (𝐵 ∈ {𝐴, 𝐶} ↔ (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
12	10, 11	sylib 122	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
13		eqcom 2233	. 2 ⊢ (𝐴 = 𝐵 ↔ 𝐵 = 𝐴)
14		eqeq2 2241	. 2 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐴 ↔ 𝐵 = 𝐶))
15	6, 12, 13, 14	oplem1 983	1 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∨ wo 715   = wceq 1397   ∈ wcel 2202  Vcvv 2802  {cpr 3670

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-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-ext 2213

This theorem depends on definitions:  df-bi 117  df-tru 1400  df-nf 1509  df-sb 1811  df-clab 2218  df-cleq 2224  df-clel 2227  df-nfc 2363  df-v 2804  df-un 3204  df-sn 3675  df-pr 3676

This theorem is referenced by:  preqr2  3852