Theorem bj-2uplth 37698

Description: The characteristic property of couples. Note that this holds without sethood hypotheses (compare opth 5463). (Contributed by BJ, 6-Oct-2018.)

Assertion

Ref	Expression
bj-2uplth	⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

Proof of Theorem bj-2uplth

Step	Hyp	Ref	Expression
1		bj-pr1eq 37679	. . . 4 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → pr1 ⦅𝐴, 𝐵⦆ = pr1 ⦅𝐶, 𝐷⦆)
2		bj-pr21val 37690	. . . 4 ⊢ pr1 ⦅𝐴, 𝐵⦆ = 𝐴
3		bj-pr21val 37690	. . . 4 ⊢ pr1 ⦅𝐶, 𝐷⦆ = 𝐶
4	1, 2, 3	3eqtr3g 2824	. . 3 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → 𝐴 = 𝐶)
5		bj-pr2eq 37693	. . . 4 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → pr2 ⦅𝐴, 𝐵⦆ = pr2 ⦅𝐶, 𝐷⦆)
6		bj-pr22val 37696	. . . 4 ⊢ pr2 ⦅𝐴, 𝐵⦆ = 𝐵
7		bj-pr22val 37696	. . . 4 ⊢ pr2 ⦅𝐶, 𝐷⦆ = 𝐷
8	5, 6, 7	3eqtr3g 2824	. . 3 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → 𝐵 = 𝐷)
9	4, 8	jca 521	. 2 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
10		bj-2upleq 37689	. . 3 ⊢ (𝐴 = 𝐶 → (𝐵 = 𝐷 → ⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆))
11	10	imp 412	. 2 ⊢ ((𝐴 = 𝐶 ∧ 𝐵 = 𝐷) → ⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆)
12	9, 11	impbii 212	1 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ ↔ (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))

This proof depends on syntax axioms:   ↔ wb 209   ∧ wa 401   = wceq 1570  pr₁ bj-cpr1 37677  ⦅bj-c2uple 37687  pr₂ bj-cpr2 37691

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-11 2195  ax-12 2216  ax-ext 2738  ax-sep 5262  ax-nul 5274  ax-pr 5409

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2745  df-cleq 2758  df-clel 2841  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5115  df-opab 5179  df-xp 5672  df-rel 5673  df-cnv 5674  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-suc 6373  df-1o 8462  df-bj-sngl 37643  df-bj-tag 37652  df-bj-proj 37668  df-bj-1upl 37675  df-bj-pr1 37678  df-bj-2upl 37688  df-bj-pr2 37692

This theorem is used by: (None)