Theorem fncofn 6684

Description: Composition of a function with domain and a function as a function with domain. Generalization of fnco 6685. (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 6667	. . . 4 ⊢ (𝐹 Fn 𝐴 → Fun 𝐹)
2		funco 6606	. . . 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 6597	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (𝐹 ∘ 𝐺) Fn dom (𝐹 ∘ 𝐺))
5		fndm 6670	. . . . . . 7 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
6	5	adantr 486	. . . . . 6 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → dom 𝐹 = 𝐴)
7	6	eqcomd 2769	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → 𝐴 = dom 𝐹)
8	7	imaeq2d 6068	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = (◡𝐺 “ dom 𝐹))
9		dmco 6273	. . . 4 ⊢ dom (𝐹 ∘ 𝐺) = (◡𝐺 “ dom 𝐹)
10	8, 9	eqtr4di 2816	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ Fun 𝐺) → (◡𝐺 “ 𝐴) = dom (𝐹 ∘ 𝐺))
11	10	fneq2d 6661	. 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 5655  dom cdm 5656   “ cima 5659   ∘ ccom 5660  Fun wfun 6558   Fn wfn 6559

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5253  ax-pr 5396

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 2102  df-mo 2567  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-id 5551  df-xp 5662  df-rel 5663  df-cnv 5664  df-co 5665  df-dm 5666  df-rn 5667  df-res 5668  df-ima 5669  df-fun 6566  df-fn 6567

This theorem is used by:  fnco  6685  fcof  6761