Theorem funresfunco 6573

Description: Composition of two functions, generalization of funco 6572. (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 6572	. 2 ⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun ((𝐹 ↾ ran 𝐺) ∘ 𝐺))
2		ssid 3953	. . . . 5 ⊢ ran 𝐺 ⊆ ran 𝐺
3		cores 6243	. . . . 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 2770	. . 3 ⊢ (𝐹 ∘ 𝐺) = ((𝐹 ↾ ran 𝐺) ∘ 𝐺)
6	5	funeqi 6552	. 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 1570   ⊆ wss 3899  ran crn 5652   ↾ cres 5653   ∘ ccom 5655  Fun wfun 6525

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 2733  ax-sep 5249  ax-pr 5391

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 2565  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  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 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-fun 6533

This theorem is used by:  fnresfnco  48055