Theorem brcog 5855

Description: Ordered pair membership in a composition. (Contributed by NM, 24-Feb-2015.)

Assertion

Ref	Expression
brcog	⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴(𝐶 ∘ 𝐷)𝐵 ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝑥,𝐶   𝑥,𝐷

Allowed substitution hints:   𝑉(𝑥)   𝑊(𝑥)

Proof of Theorem brcog

Dummy variables 𝑦 𝑧 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		breq1 5111	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦𝐷𝑥 ↔ 𝐴𝐷𝑥))
2		breq2 5112	. . . 4 ⊢ (𝑧 = 𝐵 → (𝑥𝐶𝑧 ↔ 𝑥𝐶𝐵))
3	1, 2	bi2anan9 650	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑧 = 𝐵) → ((𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧) ↔ (𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
4	3	exbidv 1956	. 2 ⊢ ((𝑦 = 𝐴 ∧ 𝑧 = 𝐵) → (∃𝑥(𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧) ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
5		df-co 5667	. 2 ⊢ (𝐶 ∘ 𝐷) = {⟨𝑦, 𝑧⟩ ∣ ∃𝑥(𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧)}
6	4, 5	brabga 5515	1 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴(𝐶 ∘ 𝐷)𝐵 ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570  ∃wex 1812   ∈ wcel 2147   class class class wbr 5108   ∘ ccom 5662

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-ext 2735  ax-sep 5255  ax-pr 5398

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-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 4841  df-br 5109  df-opab 5174  df-co 5667

This theorem is used by:  opelco2g  5856  brcogw  5857  brco  5860  brcodir  6131  predtrss  6351  brtpos2  8287  ertr  8771  relexpindlem  15259  znleval  21915  fcoinvbr  33297  opelco3  36754  brxrn  39581  eqvreltr  39889  frege124d  45022  funressnfv  48386  dfatcolem  48598