Theorem f0rn0 6659

Description: If there is no element in the range of a function, its domain must be empty. (Contributed by Alexander van der Vekens, 12-Jul-2018.)

Assertion

Ref	Expression
f0rn0	⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → 𝑋 = ∅)

Distinct variable groups:   𝑦,𝐸   𝑦,𝑌

Allowed substitution hint:   𝑋(𝑦)

Proof of Theorem f0rn0

Step	Hyp	Ref	Expression
1		fdm 6609	. . 3 ⊢ (𝐸:𝑋⟶𝑌 → dom 𝐸 = 𝑋)
2		frn 6607	. . . . . . . . 9 ⊢ (𝐸:𝑋⟶𝑌 → ran 𝐸 ⊆ 𝑌)
3		ralnex 3167	. . . . . . . . . 10 ⊢ (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 ↔ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)
4		disj 4381	. . . . . . . . . . 11 ⊢ ((𝑌 ∩ ran 𝐸) = ∅ ↔ ∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸)
5		df-ss 3904	. . . . . . . . . . . 12 ⊢ (ran 𝐸 ⊆ 𝑌 ↔ (ran 𝐸 ∩ 𝑌) = ran 𝐸)
6		incom 4135	. . . . . . . . . . . . . 14 ⊢ (ran 𝐸 ∩ 𝑌) = (𝑌 ∩ ran 𝐸)
7	6	eqeq1i 2743	. . . . . . . . . . . . 13 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 ↔ (𝑌 ∩ ran 𝐸) = ran 𝐸)
8		eqtr2 2762	. . . . . . . . . . . . . 14 ⊢ (((𝑌 ∩ ran 𝐸) = ran 𝐸 ∧ (𝑌 ∩ ran 𝐸) = ∅) → ran 𝐸 = ∅)
9	8	ex 413	. . . . . . . . . . . . 13 ⊢ ((𝑌 ∩ ran 𝐸) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
10	7, 9	sylbi 216	. . . . . . . . . . . 12 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
11	5, 10	sylbi 216	. . . . . . . . . . 11 ⊢ (ran 𝐸 ⊆ 𝑌 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
12	4, 11	syl5bir 242	. . . . . . . . . 10 ⊢ (ran 𝐸 ⊆ 𝑌 → (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
13	3, 12	syl5bir 242	. . . . . . . . 9 ⊢ (ran 𝐸 ⊆ 𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
14	2, 13	syl 17	. . . . . . . 8 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
15	14	imp 407	. . . . . . 7 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → ran 𝐸 = ∅)
16	15	adantl 482	. . . . . 6 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → ran 𝐸 = ∅)
17		dm0rn0 5834	. . . . . 6 ⊢ (dom 𝐸 = ∅ ↔ ran 𝐸 = ∅)
18	16, 17	sylibr 233	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → dom 𝐸 = ∅)
19		eqeq1 2742	. . . . . . 7 ⊢ (𝑋 = dom 𝐸 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
20	19	eqcoms 2746	. . . . . 6 ⊢ (dom 𝐸 = 𝑋 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
21	20	adantr 481	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
22	18, 21	mpbird 256	. . . 4 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → 𝑋 = ∅)
23	22	exp32 421	. . 3 ⊢ (dom 𝐸 = 𝑋 → (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅)))
24	1, 23	mpcom 38	. 2 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅))
25	24	imp 407	1 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → 𝑋 = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 205   ∧ wa 396   = wceq 1539   ∈ wcel 2106  ∀wral 3064  ∃wrex 3065   ∩ cin 3886   ⊆ wss 3887  ∅c0 4256  dom cdm 5589  ran crn 5590  ⟶wf 6429

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1798  ax-4 1812  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-11 2154  ax-12 2171  ax-ext 2709  ax-sep 5223  ax-nul 5230  ax-pr 5352

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-3an 1088  df-tru 1542  df-fal 1552  df-ex 1783  df-nf 1787  df-sb 2068  df-clab 2716  df-cleq 2730  df-clel 2816  df-ral 3069  df-rex 3070  df-rab 3073  df-v 3434  df-dif 3890  df-un 3892  df-in 3894  df-ss 3904  df-nul 4257  df-if 4460  df-sn 4562  df-pr 4564  df-op 4568  df-br 5075  df-opab 5137  df-cnv 5597  df-dm 5599  df-rn 5600  df-fn 6436  df-f 6437

This theorem is referenced by: (None)