Theorem fcomptss 45948

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 6723	. 2 ⊢ (𝜑 → 𝐹:𝐴⟶𝐶)
5		fcompt 7129	. 2 ⊢ ((𝐺:𝐶⟶𝐷 ∧ 𝐹:𝐴⟶𝐶) → (𝐺 ∘ 𝐹) = (𝑥 ∈ 𝐴 ↦ (𝐺‘(𝐹‘𝑥))))
6	1, 4, 5	syl2anc 595	1 ⊢ (𝜑 → (𝐺 ∘ 𝐹) = (𝑥 ∈ 𝐴 ↦ (𝐺‘(𝐹‘𝑥))))

This proof depends on syntax axioms:   → wi 4   = wceq 1569   ⊆ wss 3904   ↦ cmpt 5191   ∘ ccom 5664  ⟶wf 6532  ‘cfv 6536

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-10 2175  ax-11 2191  ax-12 2212  ax-ext 2734  ax-sep 5256  ax-nul 5268  ax-pr 5403

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-3an 1104  df-tru 1572  df-fal 1582  df-ex 1809  df-nf 1813  df-sb 2096  df-mo 2566  df-eu 2596  df-clab 2741  df-cleq 2754  df-clel 2837  df-nfc 2911  df-ne 2958  df-ral 3079  df-rex 3089  df-rab 3416  df-v 3456  df-sbc 3744  df-csb 3853  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4487  df-sn 4589  df-pr 4591  df-op 4595  df-uni 4872  df-br 5109  df-opab 5173  df-mpt 5192  df-id 5555  df-xp 5666  df-rel 5667  df-cnv 5668  df-co 5669  df-dm 5670  df-rn 5671  df-res 5672  df-ima 5673  df-iota 6492  df-fun 6538  df-fn 6539  df-f 6540  df-fv 6544

This theorem is used by:  ovolval2lem  47385  ovolval5lem2  47395