Theorem cocnv 38462

Description: Composition with a function and then with the converse. (Contributed by Jeff Madsen, 2-Sep-2009.)

Assertion

Ref	Expression
cocnv	⊢ ((Fun 𝐹 ∧ Fun 𝐺) → ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ↾ ran 𝐺))

Proof of Theorem cocnv

Step	Hyp	Ref	Expression
1		coass 6266	. 2 ⊢ ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ∘ (𝐺 ∘ ◡𝐺))
2		funcocnv2 6847	. . . . 5 ⊢ (Fun 𝐺 → (𝐺 ∘ ◡𝐺) = ( I ↾ ran 𝐺))
3	2	adantl 487	. . . 4 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐺 ∘ ◡𝐺) = ( I ↾ ran 𝐺))
4	3	coeq2d 5846	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ∘ ( I ↾ ran 𝐺)))
5		resco 6250	. . . 4 ⊢ ((𝐹 ∘ I ) ↾ ran 𝐺) = (𝐹 ∘ ( I ↾ ran 𝐺))
6		funrel 6554	. . . . . . 7 ⊢ (Fun 𝐹 → Rel 𝐹)
7		coi1 6263	. . . . . . 7 ⊢ (Rel 𝐹 → (𝐹 ∘ I ) = 𝐹)
8	6, 7	syl 18	. . . . . 6 ⊢ (Fun 𝐹 → (𝐹 ∘ I ) = 𝐹)
9	8	reseq1d 5975	. . . . 5 ⊢ (Fun 𝐹 → ((𝐹 ∘ I ) ↾ ran 𝐺) = (𝐹 ↾ ran 𝐺))
10	9	adantr 486	. . . 4 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → ((𝐹 ∘ I ) ↾ ran 𝐺) = (𝐹 ↾ ran 𝐺))
11	5, 10	eqtr3id 2811	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ ( I ↾ ran 𝐺)) = (𝐹 ↾ ran 𝐺))
12	4, 11	eqtrd 2797	. 2 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ↾ ran 𝐺))
13	1, 12	eqtrid 2809	1 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ↾ ran 𝐺))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1570   I cid 5553  ^◡ccnv 5658  ran crn 5660   ↾ cres 5661   ∘ ccom 5663  Rel wrel 5664  Fun wfun 6531

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-11 2194  ax-ext 2734  ax-sep 5255  ax-pr 5402

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-sb 2100  df-clab 2741  df-cleq 2754  df-clel 2837  df-ral 3079  df-rex 3089  df-rab 3415  df-v 3455  df-dif 3905  df-un 3907  df-in 3909  df-ss 3919  df-nul 4283  df-if 4486  df-sn 4588  df-pr 4590  df-op 4594  df-br 5108  df-opab 5172  df-id 5554  df-xp 5665  df-rel 5666  df-cnv 5667  df-co 5668  df-dm 5669  df-rn 5670  df-res 5671  df-fun 6539

This theorem is used by: (None)