Theorem fdifsuppconst 33379

Description: A function is a zero constant outside of its support. (Contributed by Thierry Arnoux, 22-Jun-2024.)

Hypothesis

Ref	Expression
fdifsuppconst.1	⊢ 𝐴 = (dom 𝐹 ∖ (𝐹 supp 𝑍))

Assertion

Ref	Expression
fdifsuppconst	⊢ ((Fun 𝐹 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊) → (𝐹 ↾ 𝐴) = (𝐴 × {𝑍}))

Proof of Theorem fdifsuppconst

Dummy variable 𝑥 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		funfn 6598	. . . . . 6 ⊢ (Fun 𝐹 ↔ 𝐹 Fn dom 𝐹)
2	1	biimpi 219	. . . . 5 ⊢ (Fun 𝐹 → 𝐹 Fn dom 𝐹)
3	2	ad2antrr 739	. . . 4 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → 𝐹 Fn dom 𝐹)
4		fdifsuppconst.1	. . . . 5 ⊢ 𝐴 = (dom 𝐹 ∖ (𝐹 supp 𝑍))
5		difssd 4076	. . . . 5 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (dom 𝐹 ∖ (𝐹 supp 𝑍)) ⊆ dom 𝐹)
6	4, 5	eqsstrid 3951	. . . 4 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → 𝐴 ⊆ dom 𝐹)
7	3, 6	fnssresd 6693	. . 3 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (𝐹 ↾ 𝐴) Fn 𝐴)
8		fnconstg 6800	. . . 4 ⊢ (𝑍 ∈ 𝑊 → (𝐴 × {𝑍}) Fn 𝐴)
9	8	adantl 487	. . 3 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (𝐴 × {𝑍}) Fn 𝐴)
10	3	adantr 486	. . . . 5 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → 𝐹 Fn dom 𝐹)
11		dmexg 7947	. . . . . 6 ⊢ (𝐹 ∈ 𝑉 → dom 𝐹 ∈ V)
12	11	ad3antlr 744	. . . . 5 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → dom 𝐹 ∈ V)
13		simplr 781	. . . . 5 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → 𝑍 ∈ 𝑊)
14	4	eleq2i 2856	. . . . . . 7 ⊢ (𝑥 ∈ 𝐴 ↔ 𝑥 ∈ (dom 𝐹 ∖ (𝐹 supp 𝑍)))
15	14	biimpi 219	. . . . . 6 ⊢ (𝑥 ∈ 𝐴 → 𝑥 ∈ (dom 𝐹 ∖ (𝐹 supp 𝑍)))
16	15	adantl 487	. . . . 5 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → 𝑥 ∈ (dom 𝐹 ∖ (𝐹 supp 𝑍)))
17	10, 12, 13, 16	fvdifsupp 8226	. . . 4 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → (𝐹‘𝑥) = 𝑍)
18		simpr 490	. . . . 5 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → 𝑥 ∈ 𝐴)
19	18	fvresd 6936	. . . 4 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → ((𝐹 ↾ 𝐴)‘𝑥) = (𝐹‘𝑥))
20		fvconst2g 7239	. . . . 5 ⊢ ((𝑍 ∈ 𝑊 ∧ 𝑥 ∈ 𝐴) → ((𝐴 × {𝑍})‘𝑥) = 𝑍)
21	20	adantll 727	. . . 4 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → ((𝐴 × {𝑍})‘𝑥) = 𝑍)
22	17, 19, 21	3eqtr4d 2808	. . 3 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → ((𝐹 ↾ 𝐴)‘𝑥) = ((𝐴 × {𝑍})‘𝑥))
23	7, 9, 22	eqfnfvd 7063	. 2 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (𝐹 ↾ 𝐴) = (𝐴 × {𝑍}))
24	23	3impa 1127	1 ⊢ ((Fun 𝐹 ∧ 𝐹 ∈ 𝑉 ∧ 𝑍 ∈ 𝑊) → (𝐹 ↾ 𝐴) = (𝐴 × {𝑍}))

This proof depends on syntax axioms:   → wi 4   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2147  Vcvv 3454   ∖ cdif 4048  {csn 4588   × cxp 5656  dom cdm 5658   ↾ cres 5660  Fun wfun 6560   Fn wfn 6561  ‘cfv 6566  (class class class)co 7451   supp csupp 8214

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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-rep 5238  ax-sep 5255  ax-nul 5266  ax-pr 5398  ax-un 7783

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 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-reu 3370  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-pw 4562  df-sn 4589  df-pr 4591  df-op 4841  df-uni 4873  df-iun 4958  df-br 5109  df-opab 5174  df-mpt 5193  df-id 5553  df-xp 5664  df-rel 5665  df-cnv 5666  df-co 5667  df-dm 5668  df-rn 5669  df-res 5670  df-ima 5671  df-iota 6520  df-fun 6568  df-fn 6569  df-f 6570  df-f1 6571  df-fo 6572  df-f1o 6573  df-fv 6574  df-ov 7454  df-oprab 7455  df-mpo 7456  df-supp 8215

This theorem is used by: (None)