Theorem fdifsuppconst 33161

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 6563	. . . . . 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 4084	. . . . 5 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (dom 𝐹 ∖ (𝐹 supp 𝑍)) ⊆ dom 𝐹)
6	4, 5	eqsstrid 3969	. . . 4 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → 𝐴 ⊆ dom 𝐹)
7	3, 6	fnssresd 6656	. . 3 ⊢ (((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) → (𝐹 ↾ 𝐴) Fn 𝐴)
8		fnconstg 6763	. . . 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 7898	. . . . . 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 2852	. . . . . . 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 8169	. . . 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 6898	. . . 4 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → ((𝐹 ↾ 𝐴)‘𝑥) = (𝐹‘𝑥))
20		fvconst2g 7201	. . . . 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 2805	. . 3 ⊢ ((((Fun 𝐹 ∧ 𝐹 ∈ 𝑉) ∧ 𝑍 ∈ 𝑊) ∧ 𝑥 ∈ 𝐴) → ((𝐹 ↾ 𝐴)‘𝑥) = ((𝐴 × {𝑍})‘𝑥))
23	7, 9, 22	eqfnfvd 7025	. 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 2145  Vcvv 3450   ∖ cdif 3896  {csn 4584   × cxp 5653  dom cdm 5655   ↾ cres 5657  Fun wfun 6527   Fn wfn 6528  ‘cfv 6533  (class class class)co 7413   supp csupp 8158

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-rep 5232  ax-sep 5251  ax-nul 5263  ax-pr 5398  ax-un 7736

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-reu 3366  df-rab 3413  df-v 3452  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-pw 4559  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-iun 4953  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-iota 6489  df-fun 6535  df-fn 6536  df-f 6537  df-f1 6538  df-fo 6539  df-f1o 6540  df-fv 6541  df-ov 7416  df-oprab 7417  df-mpo 7418  df-supp 8159

This theorem is used by: (None)