Theorem preqr1 3817

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 3744	. . . 4 ⊢ 𝐴 ∈ {𝐴, 𝐶}
3		eleq2 2270	. . . 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 3659	. . 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 3744	. . . 4 ⊢ 𝐵 ∈ {𝐵, 𝐶}
9		eleq2 2270	. . . 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 3659	. . 3 ⊢ (𝐵 ∈ {𝐴, 𝐶} ↔ (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
12	10, 11	sylib 122	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
13		eqcom 2208	. 2 ⊢ (𝐴 = 𝐵 ↔ 𝐵 = 𝐴)
14		eqeq2 2216	. 2 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐴 ↔ 𝐵 = 𝐶))
15	6, 12, 13, 14	oplem1 978	1 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∨ wo 710   = wceq 1373   ∈ wcel 2177  Vcvv 2773  {cpr 3639

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 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-un 3174  df-sn 3644  df-pr 3645

This theorem is referenced by:  preqr2  3818