Theorem preqr1 3695

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 3629	. . . 4 ⊢ 𝐴 ∈ {𝐴, 𝐶}
3		eleq2 2203	. . . 4 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 ∈ {𝐴, 𝐶} ↔ 𝐴 ∈ {𝐵, 𝐶}))
4	2, 3	mpbii 147	. . 3 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 ∈ {𝐵, 𝐶})
5	1	elpr 3548	. . 3 ⊢ (𝐴 ∈ {𝐵, 𝐶} ↔ (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
6	4, 5	sylib 121	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐴 = 𝐵 ∨ 𝐴 = 𝐶))
7		preqr1.2	. . . . 5 ⊢ 𝐵 ∈ V
8	7	prid1 3629	. . . 4 ⊢ 𝐵 ∈ {𝐵, 𝐶}
9		eleq2 2203	. . . 4 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 ∈ {𝐴, 𝐶} ↔ 𝐵 ∈ {𝐵, 𝐶}))
10	8, 9	mpbiri 167	. . 3 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐵 ∈ {𝐴, 𝐶})
11	7	elpr 3548	. . 3 ⊢ (𝐵 ∈ {𝐴, 𝐶} ↔ (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
12	10, 11	sylib 121	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
13		eqcom 2141	. 2 ⊢ (𝐴 = 𝐵 ↔ 𝐵 = 𝐴)
14		eqeq2 2149	. 2 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐴 ↔ 𝐵 = 𝐶))
15	6, 12, 13, 14	oplem1 959	1 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∨ wo 697   = wceq 1331   ∈ wcel 1480  Vcvv 2686  {cpr 3528

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 698  ax-5 1423  ax-7 1424  ax-gen 1425  ax-ie1 1469  ax-ie2 1470  ax-8 1482  ax-10 1483  ax-11 1484  ax-i12 1485  ax-bndl 1486  ax-4 1487  ax-17 1506  ax-i9 1510  ax-ial 1514  ax-i5r 1515  ax-ext 2121

This theorem depends on definitions:  df-bi 116  df-tru 1334  df-nf 1437  df-sb 1736  df-clab 2126  df-cleq 2132  df-clel 2135  df-nfc 2270  df-v 2688  df-un 3075  df-sn 3533  df-pr 3534

This theorem is referenced by:  preqr2  3696