Theorem fncofn 6656

Description: Composition of a function with domain and a function as a function with domain. Generalization of fnco 6657. (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 6639	. . . 4 ⊢ (𝐹 Fn 𝐴 → Fun 𝐹)
2		funco 6580	. . . 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 6571	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn dom (𝐹 ∘ 𝐺))
5		fndm 6642	. . . . . . 7 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
6	5	adantr 486	. . . . . 6 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → dom 𝐹 = 𝐴)
7	6	eqcomd 2771	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → 𝐴 = dom 𝐹)
8	7	imaeq2d 6064	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = (◡𝐺 “ dom 𝐹))
9		dmco 6258	. . . 4 ⊢ dom (𝐹 ∘ 𝐺) = (◡𝐺 “ dom 𝐹)
10	8, 9	eqtr4di 2818	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = dom (𝐹 ∘ 𝐺))
11	10	fneq2d 6633	. 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 5662  dom cdm 5663   “ cima 5666   ∘ ccom 5667  Fun wfun 6534   Fn wfn 6535

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 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2737  ax-sep 5259  ax-pr 5406

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 2569  df-eu 2599  df-clab 2744  df-cleq 2757  df-clel 2840  df-nfc 2914  df-ral 3082  df-rex 3092  df-rab 3419  df-v 3459  df-dif 3909  df-un 3911  df-in 3913  df-ss 3923  df-nul 4287  df-if 4490  df-sn 4592  df-pr 4594  df-op 4598  df-br 5112  df-opab 5176  df-id 5558  df-xp 5669  df-rel 5670  df-cnv 5671  df-co 5672  df-dm 5673  df-rn 5674  df-res 5675  df-ima 5676  df-fun 6542  df-fn 6543

This theorem is used by:  fnco  6657  fcof  6733