Theorem f0rn0 5567

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 5519	. . 3 ⊢ (𝐸:𝑋⟶𝑌 → dom 𝐸 = 𝑋)
2		frn 5522	. . . . . . . . 9 ⊢ (𝐸:𝑋⟶𝑌 → ran 𝐸 ⊆ 𝑌)
3		ralnex 2532	. . . . . . . . . 10 ⊢ (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 ↔ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)
4		disj 3561	. . . . . . . . . . 11 ⊢ ((𝑌 ∩ ran 𝐸) = ∅ ↔ ∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸)
5		df-ss 3227	. . . . . . . . . . . 12 ⊢ (ran 𝐸 ⊆ 𝑌 ↔ (ran 𝐸 ∩ 𝑌) = ran 𝐸)
6		incom 3415	. . . . . . . . . . . . . 14 ⊢ (ran 𝐸 ∩ 𝑌) = (𝑌 ∩ ran 𝐸)
7	6	eqeq1i 2242	. . . . . . . . . . . . 13 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 ↔ (𝑌 ∩ ran 𝐸) = ran 𝐸)
8		eqtr2 2253	. . . . . . . . . . . . . 14 ⊢ (((𝑌 ∩ ran 𝐸) = ran 𝐸 ∧ (𝑌 ∩ ran 𝐸) = ∅) → ran 𝐸 = ∅)
9	8	ex 115	. . . . . . . . . . . . 13 ⊢ ((𝑌 ∩ ran 𝐸) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
10	7, 9	sylbi 121	. . . . . . . . . . . 12 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
11	5, 10	sylbi 121	. . . . . . . . . . 11 ⊢ (ran 𝐸 ⊆ 𝑌 → ((𝑌 ∩ ran 𝐸) = ∅ → ran 𝐸 = ∅))
12	4, 11	biimtrrid 153	. . . . . . . . . 10 ⊢ (ran 𝐸 ⊆ 𝑌 → (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
13	3, 12	biimtrrid 153	. . . . . . . . 9 ⊢ (ran 𝐸 ⊆ 𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
14	2, 13	syl 14	. . . . . . . 8 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → ran 𝐸 = ∅))
15	14	imp 124	. . . . . . 7 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → ran 𝐸 = ∅)
16	15	adantl 277	. . . . . 6 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → ran 𝐸 = ∅)
17		dm0rn0 4978	. . . . . 6 ⊢ (dom 𝐸 = ∅ ↔ ran 𝐸 = ∅)
18	16, 17	sylibr 134	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → dom 𝐸 = ∅)
19		eqeq1 2241	. . . . . . 7 ⊢ (𝑋 = dom 𝐸 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
20	19	eqcoms 2237	. . . . . 6 ⊢ (dom 𝐸 = 𝑋 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
21	20	adantr 276	. . . . 5 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
22	18, 21	mpbird 167	. . . 4 ⊢ ((dom 𝐸 = 𝑋 ∧ (𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)) → 𝑋 = ∅)
23	22	exp32 365	. . 3 ⊢ (dom 𝐸 = 𝑋 → (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅)))
24	1, 23	mpcom 36	. 2 ⊢ (𝐸:𝑋⟶𝑌 → (¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸 → 𝑋 = ∅))
25	24	imp 124	1 ⊢ ((𝐸:𝑋⟶𝑌 ∧ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸) → 𝑋 = ∅)

Syntax hints:  ¬ wn 3   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1398   ∈ wcel 2205  ∀wral 2522  ∃wrex 2523   ∩ cin 3213   ⊆ wss 3214  ∅c0 3512  dom cdm 4754  ran crn 4755  ⟶wf 5353

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 619  ax-in2 620  ax-io 717  ax-5 1496  ax-7 1497  ax-gen 1498  ax-ie1 1542  ax-ie2 1543  ax-8 1553  ax-10 1554  ax-11 1555  ax-i12 1556  ax-bndl 1558  ax-4 1559  ax-17 1575  ax-i9 1579  ax-ial 1583  ax-i5r 1584  ax-14 2208  ax-ext 2216  ax-sep 4233  ax-pow 4292  ax-pr 4327

This theorem depends on definitions:  df-bi 117  df-3an 1007  df-tru 1401  df-fal 1404  df-nf 1510  df-sb 1812  df-eu 2085  df-mo 2086  df-clab 2221  df-cleq 2227  df-clel 2230  df-nfc 2375  df-ral 2527  df-rex 2528  df-v 2817  df-dif 3216  df-un 3218  df-in 3220  df-ss 3227  df-nul 3513  df-pw 3676  df-sn 3700  df-pr 3701  df-op 3703  df-br 4115  df-opab 4177  df-cnv 4762  df-dm 4764  df-rn 4765  df-fn 5360  df-f 5361

This theorem is referenced by: (None)