Theorem cocnv 38660

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 6267	. 2 ⊢ ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ∘ (𝐺 ∘ ◡𝐺))
2		funcocnv2 6850	. . . . 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 5839	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ∘ ( I ↾ ran 𝐺)))
5		resco 6251	. . . 4 ⊢ ((𝐹 ∘ I ) ↾ ran 𝐺) = (𝐹 ∘ ( I ↾ ran 𝐺))
6		funrel 6556	. . . . . . 7 ⊢ (Fun 𝐹 → Rel 𝐹)
7		coi1 6264	. . . . . . 7 ⊢ (Rel 𝐹 → (𝐹 ∘ I ) = 𝐹)
8	6, 7	syl 18	. . . . . 6 ⊢ (Fun 𝐹 → (𝐹 ∘ I ) = 𝐹)
9	8	reseq1d 5968	. . . . 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 2809	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ ( I ↾ ran 𝐺)) = (𝐹 ↾ ran 𝐺))
12	4, 11	eqtrd 2795	. 2 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ↾ ran 𝐺))
13	1, 12	eqtrid 2807	1 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ↾ ran 𝐺))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   = wceq 1569   I cid 5544  ^◡ccnv 5649  ran crn 5651   ↾ cres 5652   ∘ ccom 5654  Rel wrel 5655  Fun wfun 6532

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-11 2193  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-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  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 4584  df-pr 4586  df-op 4590  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 6540

This theorem is used by: (None)