Theorem fvco4 5447

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 5446	. . 3 ⊢ ((𝐾:𝐴⟶𝑋 ∧ 𝑢 ∈ 𝐴) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
2	1	ad2ant2r 498	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
3		simplr 502	. . . 4 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻 ∘ 𝐾) = 𝐹)
4	3	eqcomd 2120	. . 3 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → 𝐹 = (𝐻 ∘ 𝐾))
5	4	fveq1d 5377	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐹‘𝑢) = ((𝐻 ∘ 𝐾)‘𝑢))
6		fveq2 5375	. . 3 ⊢ (𝑥 = (𝐾‘𝑢) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
7	6	ad2antll 480	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
8	2, 5, 7	3eqtr4rd 2158	1 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐹‘𝑢))

Syntax hints:   → wi 4   ∧ wa 103   = wceq 1314   ∈ wcel 1463   ∘ ccom 4503  ⟶wf 5077  ‘cfv 5081

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 681  ax-5 1406  ax-7 1407  ax-gen 1408  ax-ie1 1452  ax-ie2 1453  ax-8 1465  ax-10 1466  ax-11 1467  ax-i12 1468  ax-bndl 1469  ax-4 1470  ax-14 1475  ax-17 1489  ax-i9 1493  ax-ial 1497  ax-i5r 1498  ax-ext 2097  ax-sep 4006  ax-pow 4058  ax-pr 4091

This theorem depends on definitions:  df-bi 116  df-3an 947  df-tru 1317  df-nf 1420  df-sb 1719  df-eu 1978  df-mo 1979  df-clab 2102  df-cleq 2108  df-clel 2111  df-nfc 2244  df-ral 2395  df-rex 2396  df-v 2659  df-sbc 2879  df-un 3041  df-in 3043  df-ss 3050  df-pw 3478  df-sn 3499  df-pr 3500  df-op 3502  df-uni 3703  df-br 3896  df-opab 3950  df-id 4175  df-xp 4505  df-rel 4506  df-cnv 4507  df-co 4508  df-dm 4509  df-rn 4510  df-res 4511  df-ima 4512  df-iota 5046  df-fun 5083  df-fn 5084  df-f 5085  df-fv 5089

This theorem is referenced by: (None)