Theorem brcogw 5811

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		3simpa 1148	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) → (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊))
2		breq2 5096	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝐴𝐷𝑥 ↔ 𝐴𝐷𝑋))
3		breq1 5095	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝑥𝐶𝐵 ↔ 𝑋𝐶𝐵))
4	2, 3	anbi12d 632	. . . . 5 ⊢ (𝑥 = 𝑋 → ((𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵) ↔ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)))
5	4	spcegv 3552	. . . 4 ⊢ (𝑋 ∈ 𝑍 → ((𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
6	5	imp 406	. . 3 ⊢ ((𝑋 ∈ 𝑍 ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
7	6	3ad2antl3 1188	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
8		brcog 5809	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴(𝐶 ∘ 𝐷)𝐵 ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
9	8	biimpar 477	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)
10	1, 7, 9	syl2an2r 685	1 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)

Syntax hints:   → wi 4   ∧ wa 395   ∧ w3a 1086   = wceq 1540  ∃wex 1779   ∈ wcel 2109   class class class wbr 5092   ∘ ccom 5623

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2701  ax-sep 5235  ax-nul 5245  ax-pr 5371

This theorem depends on definitions:  df-bi 207  df-an 396  df-or 848  df-3an 1088  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2066  df-clab 2708  df-cleq 2721  df-clel 2803  df-rab 3395  df-v 3438  df-dif 3906  df-un 3908  df-ss 3920  df-nul 4285  df-if 4477  df-sn 4578  df-pr 4580  df-op 4584  df-br 5093  df-opab 5155  df-co 5628

This theorem is referenced by:  utop2nei  24136  utop3cls  24137  iunrelexpuztr  43702  frege96d  43732  frege98d  43736