Theorem brcog 5845

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 5105	. . . 4 ⊢ (𝑦 = 𝐴 → (𝑦𝐷𝑥 ↔ 𝐴𝐷𝑥))
2		breq2 5106	. . . 4 ⊢ (𝑧 = 𝐵 → (𝑥𝐶𝑧 ↔ 𝑥𝐶𝐵))
3	1, 2	bi2anan9 650	. . 3 ⊢ ((𝑦 = 𝐴 ∧ 𝑧 = 𝐵) → ((𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧) ↔ (𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
4	3	exbidv 1953	. 2 ⊢ ((𝑦 = 𝐴 ∧ 𝑧 = 𝐵) → (∃𝑥(𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧) ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
5		df-co 5660	. 2 ⊢ (𝐶 ∘ 𝐷) = {⟨𝑦, 𝑧⟩ ∣ ∃𝑥(𝑦𝐷𝑥 ∧ 𝑥𝐶𝑧)}
6	4, 5	brabga 5508	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 1569  ∃wex 1811   ∈ wcel 2144   class class class wbr 5102   ∘ ccom 5655

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732  ax-sep 5248  ax-pr 5391

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-rab 3413  df-v 3452  df-ss 3897  df-dif 4046  df-in 4090  df-un 4150  df-nul 4278  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-co 5660

This theorem is used by:  opelco2g  5846  brcogw  5847  brco  5849  brcodir  6115  predtrss  6328  brtpos2  8257  ertr  8741  relexpindlem  15225  znleval  21869  fcoinvbr  33250  opelco3  36663  brxrn  39452  eqvreltr  39760  frege124d  44898  funressnfv  48249  dfatcolem  48461