Theorem brcogw 4810

Description: Ordered pair membership in a composition. (Contributed by Thierry Arnoux, 14-Jan-2018.)

Assertion

Ref	Expression
brcogw	⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)

Proof of Theorem brcogw

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		simpl1 1001	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐴 ∈ 𝑉)
2		simpl2 1002	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐵 ∈ 𝑊)
3		breq2 4021	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝐴𝐷𝑥 ↔ 𝐴𝐷𝑋))
4		breq1 4020	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝑥𝐶𝐵 ↔ 𝑋𝐶𝐵))
5	3, 4	anbi12d 473	. . . . 5 ⊢ (𝑥 = 𝑋 → ((𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵) ↔ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)))
6	5	spcegv 2839	. . . 4 ⊢ (𝑋 ∈ 𝑍 → ((𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
7	6	imp 124	. . 3 ⊢ ((𝑋 ∈ 𝑍 ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
8	7	3ad2antl3 1162	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
9		brcog 4808	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴(𝐶 ∘ 𝐷)𝐵 ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
10	9	biimpar 297	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)
11	1, 2, 8, 10	syl21anc 1247	1 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)

Syntax hints:   → wi 4   ∧ wa 104   ∧ w3a 979   = wceq 1363  ∃wex 1502   ∈ wcel 2159   class class class wbr 4017   ∘ ccom 4644

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1457  ax-7 1458  ax-gen 1459  ax-ie1 1503  ax-ie2 1504  ax-8 1514  ax-10 1515  ax-11 1516  ax-i12 1517  ax-bndl 1519  ax-4 1520  ax-17 1536  ax-i9 1540  ax-ial 1544  ax-i5r 1545  ax-14 2162  ax-ext 2170  ax-sep 4135  ax-pow 4188  ax-pr 4223

This theorem depends on definitions:  df-bi 117  df-3an 981  df-tru 1366  df-nf 1471  df-sb 1773  df-eu 2040  df-mo 2041  df-clab 2175  df-cleq 2181  df-clel 2184  df-nfc 2320  df-v 2753  df-un 3147  df-in 3149  df-ss 3156  df-pw 3591  df-sn 3612  df-pr 3613  df-op 3615  df-br 4018  df-opab 4079  df-co 4649

This theorem is referenced by: (None)