Theorem funresfunco 6582

Description: Composition of two functions, generalization of funco 6581. (Contributed by Alexander van der Vekens, 25-Jul-2017.)

Assertion

Ref	Expression
funresfunco	⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun (𝐹 ∘ 𝐺))

Proof of Theorem funresfunco

Step	Hyp	Ref	Expression
1		funco 6581	. 2 ⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun ((𝐹 ↾ ran 𝐺) ∘ 𝐺))
2		ssid 3930	. . . . 5 ⊢ ran 𝐺 ⊆ ran 𝐺
3		cores 6250	. . . . 5 ⊢ (ran 𝐺 ⊆ ran 𝐺 → ((𝐹 ↾ ran 𝐺) ∘ 𝐺) = (𝐹 ∘ 𝐺))
4	2, 3	ax-mp 5	. . . 4 ⊢ ((𝐹 ↾ ran 𝐺) ∘ 𝐺) = (𝐹 ∘ 𝐺)
5	4	eqcomi 2769	. . 3 ⊢ (𝐹 ∘ 𝐺) = ((𝐹 ↾ ran 𝐺) ∘ 𝐺)
6	5	funeqi 6561	. 2 ⊢ (Fun (𝐹 ∘ 𝐺) ↔ Fun ((𝐹 ↾ ran 𝐺) ∘ 𝐺))
7	1, 6	sylibr 237	1 ⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun (𝐹 ∘ 𝐺))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   ⊆ wss 3895  ran crn 5651   ↾ cres 5652   ∘ ccom 5654  Fun wfun 6533

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-fun 6541

This theorem is used by:  fnresfnco  48205