Theorem f0rn0 5585

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 5537	. . 3 ⊢ (𝐸:𝑋⟶𝑌 → dom 𝐸 = 𝑋)
2		frn 5540	. . . . . . . . 9 ⊢ (𝐸:𝑋⟶𝑌 → ran 𝐸 ⊆ 𝑌)
3		ralnex 2538	. . . . . . . . . 10 ⊢ (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 ↔ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)
4		disj 3573	. . . . . . . . . . 11 ⊢ ((𝑌 ∩ ran 𝐸) = ∅ ↔ ∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸)
5		df-ss 3233	. . . . . . . . . . . 12 ⊢ (ran 𝐸 ⊆ 𝑌 ↔ (ran 𝐸 ∩ 𝑌) = ran 𝐸)
6		incom 3421	. . . . . . . . . . . . . 14 ⊢ (ran 𝐸 ∩ 𝑌) = (𝑌 ∩ ran 𝐸)
7	6	eqeq1i 2246	. . . . . . . . . . . . 13 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 ↔ (𝑌 ∩ ran 𝐸) = ran 𝐸)
8		eqtr2 2257	. . . . . . . . . . . . . 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 4996	. . . . . 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 2245	. . . . . . 7 ⊢ (𝑋 = dom 𝐸 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
20	19	eqcoms 2241	. . . . . 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 1402   ∈ wcel 2209  ∀wral 2528  ∃wrex 2529   ∩ cin 3219   ⊆ wss 3220  ∅c0 3520  dom cdm 4772  ran crn 4773  ⟶wf 5371

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 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-sep 4247  ax-pow 4309  ax-pr 4344

This theorem depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ral 2533  df-rex 2534  df-v 2823  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-nul 3521  df-pw 3690  df-sn 3714  df-pr 3715  df-op 3717  df-br 4129  df-opab 4191  df-cnv 4780  df-dm 4782  df-rn 4783  df-fn 5378  df-f 5379

This theorem is referenced by: (None)