Theorem preqr1 3770

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 3700	. . . 4 ⊢ 𝐴 ∈ {𝐴, 𝐶}
3		eleq2 2241	. . . 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 3615	. . 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 3700	. . . 4 ⊢ 𝐵 ∈ {𝐵, 𝐶}
9		eleq2 2241	. . . 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 3615	. . 3 ⊢ (𝐵 ∈ {𝐴, 𝐶} ↔ (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
12	10, 11	sylib 122	. 2 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → (𝐵 = 𝐴 ∨ 𝐵 = 𝐶))
13		eqcom 2179	. 2 ⊢ (𝐴 = 𝐵 ↔ 𝐵 = 𝐴)
14		eqeq2 2187	. 2 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐴 ↔ 𝐵 = 𝐶))
15	6, 12, 13, 14	oplem1 975	1 ⊢ ({𝐴, 𝐶} = {𝐵, 𝐶} → 𝐴 = 𝐵)

Syntax hints:   → wi 4   ∨ wo 708   = wceq 1353   ∈ wcel 2148  Vcvv 2739  {cpr 3595

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-un 3135  df-sn 3600  df-pr 3601

This theorem is referenced by:  preqr2  3771