Theorem snopeqop 4969

Description: Equivalence for an ordered pair equal to a singleton of an ordered pair. (Contributed by AV, 18-Sep-2020.)

Hypotheses

Ref	Expression
snopeqop.a	⊢ 𝐴 ∈ V
snopeqop.b	⊢ 𝐵 ∈ V
snopeqop.c	⊢ 𝐶 ∈ V
snopeqop.d	⊢ 𝐷 ∈ V

Assertion

Ref	Expression
snopeqop	⊢ ({⟨𝐴, 𝐵⟩} = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ∧ 𝐶 = {𝐴}))

Proof of Theorem snopeqop

Step	Hyp	Ref	Expression
1		snopeqop.a	. . . 4 ⊢ 𝐴 ∈ V
2		snopeqop.b	. . . 4 ⊢ 𝐵 ∈ V
3		snopeqop.c	. . . 4 ⊢ 𝐶 ∈ V
4	1, 2, 3	opeqsn 4967	. . 3 ⊢ (⟨𝐴, 𝐵⟩ = {𝐶} ↔ (𝐴 = 𝐵 ∧ 𝐶 = {𝐴}))
5	4	anbi2i 730	. 2 ⊢ ((𝐶 = 𝐷 ∧ ⟨𝐴, 𝐵⟩ = {𝐶}) ↔ (𝐶 = 𝐷 ∧ (𝐴 = 𝐵 ∧ 𝐶 = {𝐴})))
6		eqcom 2629	. . 3 ⊢ ({⟨𝐴, 𝐵⟩} = ⟨𝐶, 𝐷⟩ ↔ ⟨𝐶, 𝐷⟩ = {⟨𝐴, 𝐵⟩})
7		snopeqop.d	. . . 4 ⊢ 𝐷 ∈ V
8		opex 4932	. . . 4 ⊢ ⟨𝐴, 𝐵⟩ ∈ V
9	3, 7, 8	opeqsn 4967	. . 3 ⊢ (⟨𝐶, 𝐷⟩ = {⟨𝐴, 𝐵⟩} ↔ (𝐶 = 𝐷 ∧ ⟨𝐴, 𝐵⟩ = {𝐶}))
10	6, 9	bitri 264	. 2 ⊢ ({⟨𝐴, 𝐵⟩} = ⟨𝐶, 𝐷⟩ ↔ (𝐶 = 𝐷 ∧ ⟨𝐴, 𝐵⟩ = {𝐶}))
11		3anan12 1051	. 2 ⊢ ((𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ∧ 𝐶 = {𝐴}) ↔ (𝐶 = 𝐷 ∧ (𝐴 = 𝐵 ∧ 𝐶 = {𝐴})))
12	5, 10, 11	3bitr4i 292	1 ⊢ ({⟨𝐴, 𝐵⟩} = ⟨𝐶, 𝐷⟩ ↔ (𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ∧ 𝐶 = {𝐴}))

Syntax hints:   ↔ wb 196   ∧ wa 384   ∧ w3a 1037   = wceq 1483   ∈ wcel 1990  Vcvv 3200  {csn 4177  ⟨cop 4183

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1722  ax-4 1737  ax-5 1839  ax-6 1888  ax-7 1935  ax-9 1999  ax-10 2019  ax-11 2034  ax-12 2047  ax-13 2246  ax-ext 2602  ax-sep 4781  ax-nul 4789  ax-pr 4906

This theorem depends on definitions:  df-bi 197  df-or 385  df-an 386  df-3an 1039  df-tru 1486  df-ex 1705  df-nf 1710  df-sb 1881  df-clab 2609  df-cleq 2615  df-clel 2618  df-nfc 2753  df-v 3202  df-dif 3577  df-un 3579  df-in 3581  df-ss 3588  df-nul 3916  df-if 4087  df-sn 4178  df-pr 4180  df-op 4184

This theorem is referenced by:  funopsn  6413  funsneqopsn  6417