Theorem fcomptss 46138

Description: Express composition of two functions as a maps-to applying both in sequence. (Contributed by Glauco Siliprandi, 3-Mar-2021.)

Hypotheses

Ref	Expression
fcomptss.a	⊢ (𝜑 → 𝐹:𝐴⟶𝐵)
fcomptss.b	⊢ (𝜑 → 𝐵 ⊆ 𝐶)
fcomptss.g	⊢ (𝜑 → 𝐺:𝐶⟶𝐷)

Assertion

Ref	Expression
fcomptss	⊢ (𝜑 → (𝐺 ∘ 𝐹) = (𝑥 ∈ 𝐴 ↦ (𝐺‘(𝐹‘𝑥))))

Distinct variable groups:   𝑥,𝐴   𝑥,𝐶   𝑥,𝐷   𝑥,𝐹   𝑥,𝐺

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem fcomptss

Step	Hyp	Ref	Expression
1		fcomptss.g	. 2 ⊢ (𝜑 → 𝐺:𝐶⟶𝐷)
2		fcomptss.a	. . 3 ⊢ (𝜑 → 𝐹:𝐴⟶𝐵)
3		fcomptss.b	. . 3 ⊢ (𝜑 → 𝐵 ⊆ 𝐶)
4	2, 3	fssd 6715	. 2 ⊢ (𝜑 → 𝐹:𝐴⟶𝐶)
5		fcompt 7122	. 2 ⊢ ((𝐺:𝐶⟶𝐷 ∧ 𝐹:𝐴⟶𝐶) → (𝐺 ∘ 𝐹) = (𝑥 ∈ 𝐴 ↦ (𝐺‘(𝐹‘𝑥))))
6	1, 4, 5	syl2anc 596	1 ⊢ (𝜑 → (𝐺 ∘ 𝐹) = (𝑥 ∈ 𝐴 ↦ (𝐺‘(𝐹‘𝑥))))

This proof depends on syntax axioms:   → wi 4   = wceq 1570   ⊆ wss 3898   ↦ cmpt 5185   ∘ ccom 5651  ⟶wf 6523  ‘cfv 6527

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 1943  ax-6 2000  ax-7 2041  ax-8 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2732  ax-sep 5248  ax-nul 5259  ax-pr 5390

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-nf 1817  df-sb 2100  df-mo 2564  df-eu 2594  df-clab 2739  df-cleq 2752  df-clel 2835  df-nfc 2909  df-ne 2956  df-ral 3077  df-rex 3087  df-rab 3413  df-v 3452  df-sbc 3739  df-csb 3847  df-dif 3901  df-un 3903  df-in 3905  df-ss 3915  df-nul 4279  df-if 4482  df-sn 4584  df-pr 4586  df-op 4590  df-uni 4867  df-br 5103  df-opab 5167  df-mpt 5186  df-id 5542  df-xp 5653  df-rel 5654  df-cnv 5655  df-co 5656  df-dm 5657  df-rn 5658  df-res 5659  df-ima 5660  df-iota 6483  df-fun 6529  df-fn 6530  df-f 6531  df-fv 6535

This theorem is used by:  ovolval2lem  47575  ovolval5lem2  47585