Theorem funresfunco 6604

Description: Composition of two functions, generalization of funco 6603. (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 6603	. 2 ⊢ ((Fun (𝐹 ↾ ran 𝐺) ∧ Fun 𝐺) → Fun ((𝐹 ↾ ran 𝐺) ∘ 𝐺))
2		ssid 3935	. . . . 5 ⊢ ran 𝐺 ⊆ ran 𝐺
3		cores 6264	. . . . 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 2772	. . 3 ⊢ (𝐹 ∘ 𝐺) = ((𝐹 ↾ ran 𝐺) ∘ 𝐺)
6	5	funeqi 6583	. 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 3900  ran crn 5657   ↾ cres 5658   ∘ ccom 5660  Fun wfun 6555

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-fun 6563

This theorem is used by:  fnresfnco  48326