Theorem fvco4 5771

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 5770	. . 3 ⊢ ((𝐾:𝐴⟶𝑋 ∧ 𝑢 ∈ 𝐴) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
2	1	ad2ant2r 513	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → ((𝐻 ∘ 𝐾)‘𝑢) = (𝐻‘(𝐾‘𝑢)))
3		simplr 533	. . . 4 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻 ∘ 𝐾) = 𝐹)
4	3	eqcomd 2244	. . 3 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → 𝐹 = (𝐻 ∘ 𝐾))
5	4	fveq1d 5692	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐹‘𝑢) = ((𝐻 ∘ 𝐾)‘𝑢))
6		fveq2 5690	. . 3 ⊢ (𝑥 = (𝐾‘𝑢) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
7	6	ad2antll 495	. 2 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐻‘(𝐾‘𝑢)))
8	2, 5, 7	3eqtr4rd 2282	1 ⊢ (((𝐾:𝐴⟶𝑋 ∧ (𝐻 ∘ 𝐾) = 𝐹) ∧ (𝑢 ∈ 𝐴 ∧ 𝑥 = (𝐾‘𝑢))) → (𝐻‘𝑥) = (𝐹‘𝑢))

Syntax hints:   → wi 4   ∧ wa 104   = wceq 1402   ∈ wcel 2209   ∘ ccom 4773  ⟶wf 5368  ‘cfv 5372

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 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4244  ax-pow 4306  ax-pr 4341

This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-sbc 3052  df-un 3224  df-in 3226  df-ss 3233  df-pw 3687  df-sn 3711  df-pr 3712  df-op 3714  df-uni 3931  df-br 4126  df-opab 4188  df-id 4433  df-xp 4775  df-rel 4776  df-cnv 4777  df-co 4778  df-dm 4779  df-rn 4780  df-res 4781  df-ima 4782  df-iota 5332  df-fun 5374  df-fn 5375  df-f 5376  df-fv 5380

This theorem is referenced by: (None)