Theorem fvco4 5586

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 5585	. . 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 2183	. . 3 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → 𝐹 = (𝐻 ∘ 𝐾))
5	4	fveq1d 5515	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐹‘𝑢) = ((𝐻 ∘ 𝐾)‘𝑢))
6		fveq2 5513	. . 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 2221	1 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐹‘𝑢))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1353   ∈ wcel 2148   ∘ ccom 4629  ⟶wf 5210  ‘cfv 5214

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4120  ax-pow 4173  ax-pr 4208

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2739  df-sbc 2963  df-un 3133  df-in 3135  df-ss 3142  df-pw 3577  df-sn 3598  df-pr 3599  df-op 3601  df-uni 3810  df-br 4003  df-opab 4064  df-id 4292  df-xp 4631  df-rel 4632  df-cnv 4633  df-co 4634  df-dm 4635  df-rn 4636  df-res 4637  df-ima 4638  df-iota 5176  df-fun 5216  df-fn 5217  df-f 5218  df-fv 5222

This theorem is referenced by: (None)