Theorem bj-2uplth 38125

Description: The characteristic property of couples. This holds without sethood hypotheses (compare opth 5450). (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 38106	. . . 4 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → pr1 ⦅𝐴, 𝐵⦆ = pr1 ⦅𝐶, 𝐷⦆)
2		bj-pr21val 38117	. . . 4 ⊢ pr1 ⦅𝐴, 𝐵⦆ = 𝐴
3		bj-pr21val 38117	. . . 4 ⊢ pr1 ⦅𝐶, 𝐷⦆ = 𝐶
4	1, 2, 3	3eqtr3g 2821	. . 3 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → 𝐴 = 𝐶)
5		bj-pr2eq 38120	. . . 4 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → pr2 ⦅𝐴, 𝐵⦆ = pr2 ⦅𝐶, 𝐷⦆)
6		bj-pr22val 38123	. . . 4 ⊢ pr2 ⦅𝐴, 𝐵⦆ = 𝐵
7		bj-pr22val 38123	. . . 4 ⊢ pr2 ⦅𝐶, 𝐷⦆ = 𝐷
8	5, 6, 7	3eqtr3g 2821	. . 3 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → 𝐵 = 𝐷)
9	4, 8	jca 521	. 2 ⊢ (⦅𝐴, 𝐵⦆ = ⦅𝐶, 𝐷⦆ → (𝐴 = 𝐶 ∧ 𝐵 = 𝐷))
10		bj-2upleq 38116	. . 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 38104  ⦅bj-c2uple 38114  pr₂ bj-cpr2 38118

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5253  ax-nul 5264  ax-pr 5396

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-xp 5662  df-rel 5663  df-cnv 5664  df-dm 5666  df-rn 5667  df-res 5668  df-ima 5669  df-suc 6389  df-1o 8507  df-bj-sngl 38070  df-bj-tag 38079  df-bj-proj 38095  df-bj-1upl 38102  df-bj-pr1 38105  df-bj-2upl 38115  df-bj-pr2 38119

This theorem is used by: (None)