Theorem fvco4 5629

Description: Value of a composition. (Contributed by BJ, 7-Jul-2022.)

Assertion

Ref	Expression
fvco4	⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐹‘𝑢))

Proof of Theorem fvco4

Step	Hyp	Ref	Expression
1		fvco3 5628	. . 3 ⊢ ((𝐾:𝐴⟶𝑋 ∧ 𝑢 ∈ 𝐴) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
2	1	ad2ant2r 509	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
3		simplr 528	. . . 4 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻 ∘ 𝐾) = 𝐹)
4	3	eqcomd 2199	. . 3 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → 𝐹 = (𝐻 ∘ 𝐾))
5	4	fveq1d 5556	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐹‘𝑢) = ((𝐻 ∘ 𝐾)‘𝑢))
6		fveq2 5554	. . 3 ⊢ (𝑥 = (𝐾‘𝑢) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
7	6	ad2antll 491	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
8	2, 5, 7	3eqtr4rd 2237	1 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐹‘𝑢))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1364   ∈ wcel 2164   ∘ ccom 4663  ⟶wf 5250  ‘cfv 5254

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-14 2167  ax-ext 2175  ax-sep 4147  ax-pow 4203  ax-pr 4238

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1472  df-sb 1774  df-eu 2045  df-mo 2046  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ral 2477  df-rex 2478  df-v 2762  df-sbc 2986  df-un 3157  df-in 3159  df-ss 3166  df-pw 3603  df-sn 3624  df-pr 3625  df-op 3627  df-uni 3836  df-br 4030  df-opab 4091  df-id 4324  df-xp 4665  df-rel 4666  df-cnv 4667  df-co 4668  df-dm 4669  df-rn 4670  df-res 4671  df-ima 4672  df-iota 5215  df-fun 5256  df-fn 5257  df-f 5258  df-fv 5262

This theorem is referenced by: (None)