Theorem fncofn 6649

Description: Composition of a function with domain and a function as a function with domain. Generalization of fnco 6650. (Contributed by AV, 17-Sep-2024.)

Assertion

Ref	Expression
fncofn	⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn (◡𝐺 “ 𝐴))

Proof of Theorem fncofn

Step	Hyp	Ref	Expression
1		fnfun 6632	. . . 4 ⊢ (𝐹 Fn 𝐴 → Fun 𝐹)
2		funco 6573	. . . 4 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → Fun (𝐹 ∘ 𝐺))
3	1, 2	sylan 592	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → Fun (𝐹 ∘ 𝐺))
4	3	funfnd 6564	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn dom (𝐹 ∘ 𝐺))
5		fndm 6635	. . . . . . 7 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
6	5	adantr 486	. . . . . 6 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → dom 𝐹 = 𝐴)
7	6	eqcomd 2766	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → 𝐴 = dom 𝐹)
8	7	imaeq2d 6056	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = (◡𝐺 “ dom 𝐹))
9		dmco 6251	. . . 4 ⊢ dom (𝐹 ∘ 𝐺) = (◡𝐺 “ dom 𝐹)
10	8, 9	eqtr4di 2813	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = dom (𝐹 ∘ 𝐺))
11	10	fneq2d 6626	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → ((𝐹 ∘ 𝐺) Fn (◡𝐺 “ 𝐴) ↔ (𝐹 ∘ 𝐺) Fn dom (𝐹 ∘ 𝐺)))
12	4, 11	mpbird 260	1 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn (◡𝐺 “ 𝐴))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570  ^◡ccnv 5654  dom cdm 5655   “ cima 5658   ∘ ccom 5659  Fun wfun 6527   Fn wfn 6528

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732  ax-sep 5251  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-nf 1817  df-sb 2100  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-br 5104  df-opab 5168  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-fun 6535  df-fn 6536

This theorem is used by:  fnco  6650  fcof  6726