Theorem f0rn0 6792

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 6744	. . 3 ⊢ (𝐸:𝑋⟶𝑌 → dom 𝐸 = 𝑋)
2		frn 6742	. . . . . . . . 9 ⊢ (𝐸:𝑋⟶𝑌 → ran 𝐸 ⊆ 𝑌)
3		ralnex 3092	. . . . . . . . . 10 ⊢ (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 ↔ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)
4		disj 4406	. . . . . . . . . . 11 ⊢ ((𝑌 ∩ ran 𝐸) = ∅ ↔ ∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸)
5		dfss2 4097	. . . . . . . . . . . 12 ⊢ (ran 𝐸 ⊆ 𝑌 ↔ (ran 𝐸 ∩ 𝑌) = ran 𝐸)
6		incom 4109	. . . . . . . . . . . . . 14 ⊢ (ran 𝐸 ∩ 𝑌) = (𝑌 ∩ ran 𝐸)
7	6	eqeq1i 2768	. . . . . . . . . . . . 13 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 ↔ (𝑌 ∩ ran 𝐸) = ran 𝐸)
8		eqtr2 2784	. . . . . . . . . . . . . 14 ⊢ (((𝑌 ∩ ran 𝐸) = ran 𝐸 ∧ (𝑌 ∩ ran 𝐸) = ∅) → ran 𝐸 = ∅)
9	8	ex 418	. . . . . . . . . . . . 13 ⊢ ((𝑌 ∩ ran 𝐸) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
10	7, 9	sylbi 220	. . . . . . . . . . . 12 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
11	5, 10	sylbi 220	. . . . . . . . . . 11 ⊢ (ran 𝐸 ⊆ 𝑌 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
12	4, 11	biimtrrid 246	. . . . . . . . . 10 ⊢ (ran 𝐸 ⊆ 𝑌 → (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
13	3, 12	biimtrrid 246	. . . . . . . . 9 ⊢ (ran 𝐸 ⊆ 𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
14	2, 13	syl 18	. . . . . . . 8 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
15	14	imp 412	. . . . . . 7 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → ran 𝐸 = ∅)
16	15	adantl 487	. . . . . 6 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → ran 𝐸 = ∅)
17		dm0rn0 5914	. . . . . 6 ⊢ (dom 𝐸 = ∅ ↔ ran 𝐸 = ∅)
18	16, 17	sylibr 237	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → dom 𝐸 = ∅)
19		eqeq1 2767	. . . . . . 7 ⊢ (𝑋 = dom 𝐸 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
20	19	eqcoms 2771	. . . . . 6 ⊢ (dom 𝐸 = 𝑋 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
21	20	adantr 486	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
22	18, 21	mpbird 260	. . . 4 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → 𝑋 = ∅)
23	22	exp32 426	. . 3 ⊢ (dom 𝐸 = 𝑋 → (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅)))
24	1, 23	mpcom 39	. 2 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅))
25	24	imp 412	1 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → 𝑋 = ∅)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2147  ∀wral 3080  ∃wrex 3090   ⊆ wss 3900   ∩ cin 4092  ∅c0 4281  dom cdm 5656  ran crn 5657  ⟶wf 6557

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-ext 2735  ax-sep 5253  ax-pr 5396

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-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-ss 3901  df-dif 4050  df-in 4094  df-un 4154  df-nul 4282  df-if 4486  df-sn 4589  df-pr 4591  df-op 4839  df-br 5107  df-opab 5172  df-cnv 5664  df-dm 5666  df-rn 5667  df-fn 6564  df-f 6565

This theorem is used by: (None)