Theorem funresfunco 6584

Description: Composition of two functions, generalization of funco 6583. (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 6583	. 2 ⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun ((𝐹 ↾ ran 𝐺) ∘ 𝐺))
2		ssid 3962	. . . . 5 ⊢ ran 𝐺 ⊆ ran 𝐺
3		cores 6255	. . . . 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 2775	. . 3 ⊢ (𝐹 ∘ 𝐺) = ((𝐹 ↾ ran 𝐺) ∘ 𝐺)
6	5	funeqi 6564	. 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 3908  ran crn 5667   ↾ cres 5668   ∘ ccom 5670  Fun wfun 6537

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 2738  ax-sep 5262  ax-pr 5409

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 2570  df-eu 2600  df-clab 2745  df-cleq 2758  df-clel 2841  df-nfc 2915  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-br 5115  df-opab 5179  df-id 5561  df-xp 5672  df-rel 5673  df-cnv 5674  df-co 5675  df-dm 5676  df-rn 5677  df-res 5678  df-fun 6545

This theorem is used by:  fnresfnco  47819