Theorem cocnv 38850

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 6281	. 2 ⊢ ((𝐹 ∘ 𝐺) ∘ ◡𝐺) = (𝐹 ∘ (𝐺 ∘ ◡𝐺))
2		funcocnv2 6869	. . . . 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 5845	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ∘ ( I ↾ ran 𝐺)))
5		resco 6263	. . . 4 ⊢ ((𝐹 ∘ I ) ↾ ran 𝐺) = (𝐹 ∘ ( I ↾ ran 𝐺))
6		funrel 6573	. . . . . . 7 ⊢ (Fun 𝐹 → Rel 𝐹)
7		coi1 6278	. . . . . . 7 ⊢ (Rel 𝐹 → (𝐹 ∘ I ) = 𝐹)
8	6, 7	syl 18	. . . . . 6 ⊢ (Fun 𝐹 → (𝐹 ∘ I ) = 𝐹)
9	8	reseq1d 5979	. . . . 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 2812	. . 3 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ ( I ↾ ran 𝐺)) = (𝐹 ↾ ran 𝐺))
12	4, 11	eqtrd 2798	. 2 ⊢ ((Fun 𝐹 ∧ Fun 𝐺) → (𝐹 ∘ (𝐺 ∘ ◡𝐺)) = (𝐹 ↾ ran 𝐺))
13	1, 12	eqtrid 2810	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 5549  ^◡ccnv 5654  ran crn 5656   ↾ cres 5657   ∘ ccom 5659  Rel wrel 5660  Fun wfun 6549

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-ext 2735  ax-sep 5252  ax-pr 5395

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 2102  df-clab 2742  df-cleq 2755  df-clel 2839  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 5171  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-fun 6557

This theorem is used by: (None)