Theorem fncofn 6657

Description: Composition of a function with domain and a function as a function with domain. Generalization of fnco 6658. (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 6640	. . . 4 ⊢ (𝐹 Fn 𝐴 → Fun 𝐹)
2		funco 6581	. . . 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 6572	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn dom (𝐹 ∘ 𝐺))
5		fndm 6643	. . . . . . 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 6051	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = (◡𝐺 “ dom 𝐹))
9		dmco 6256	. . . 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 6634	. 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 1569  ^◡ccnv 5649  dom cdm 5650   “ cima 5653   ∘ ccom 5654  Fun wfun 6533   Fn wfn 6534

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-10 2177  ax-11 2193  ax-12 2212  ax-ext 2732  ax-sep 5248  ax-pr 5390

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-nf 1816  df-sb 2099  df-mo 2564  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-ss 3896  df-dif 4047  df-in 4091  df-un 4151  df-nul 4279  df-if 4482  df-sn 4585  df-pr 4587  df-op 4835  df-br 5103  df-opab 5167  df-id 5545  df-xp 5656  df-rel 5657  df-cnv 5658  df-co 5659  df-dm 5660  df-rn 5661  df-res 5662  df-ima 5663  df-fun 6541  df-fn 6542

This theorem is used by:  fnco  6658  fcof  6734