Theorem brcogw 5854

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 1166	. 2 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) → (𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊))
2		breq2 5113	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝐴𝐷𝑥 ↔ 𝐴𝐷𝑋))
3		breq1 5112	. . . . . 6 ⊢ (𝑥 = 𝑋 → (𝑥𝐶𝐵 ↔ 𝑋𝐶𝐵))
4	2, 3	anbi12d 643	. . . . 5 ⊢ (𝑥 = 𝑋 → ((𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵) ↔ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)))
5	4	spcegv 3556	. . . 4 ⊢ (𝑋 ∈ 𝑍 → ((𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
6	5	imp 411	. . 3 ⊢ ((𝑋 ∈ 𝑍 ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
7	6	3ad2antl3 1206	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵))
8		brcog 5852	. . 3 ⊢ ((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) → (𝐴(𝐶 ∘ 𝐷)𝐵 ↔ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)))
9	8	biimpar 482	. 2 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊) ∧ ∃𝑥(𝐴𝐷𝑥 ∧ 𝑥𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)
10	1, 7, 9	syl2an2r 697	1 ⊢ (((𝐴 ∈ 𝑉 ∧ 𝐵 ∈ 𝑊 ∧ 𝑋 ∈ 𝑍) ∧ (𝐴𝐷𝑋 ∧ 𝑋𝐶𝐵)) → 𝐴(𝐶 ∘ 𝐷)𝐵)

This proof depends on syntax axioms:   → wi 4   ∧ wa 400   ∧ w3a 1103   = wceq 1570  ∃wex 1809   ∈ wcel 2143   class class class wbr 5109   ∘ ccom 5665

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735  ax-sep 5257  ax-pr 5404

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1105  df-tru 1573  df-fal 1583  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-rab 3417  df-v 3457  df-dif 3908  df-un 3910  df-in 3912  df-ss 3922  df-nul 4287  df-if 4488  df-sn 4590  df-pr 4592  df-op 4596  df-br 5110  df-opab 5174  df-co 5670

This theorem is used by:  utop2nei  24416  utop3cls  24417  iunrelexpuztr  44473  frege96d  44503  frege98d  44507