Theorem f0rn0 5412

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 5373	. . 3 ⊢ (𝐸:𝑋⟶𝑌 → dom 𝐸 = 𝑋)
2		frn 5376	. . . . . . . . 9 ⊢ (𝐸:𝑋⟶𝑌 → ran 𝐸 ⊆ 𝑌)
3		ralnex 2465	. . . . . . . . . 10 ⊢ (∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸 ↔ ¬ ∃𝑦 ∈ 𝑌 𝑦 ∈ ran 𝐸)
4		disj 3473	. . . . . . . . . . 11 ⊢ ((𝑌 ∩ ran 𝐸) = ∅ ↔ ∀𝑦 ∈ 𝑌 ¬ 𝑦 ∈ ran 𝐸)
5		df-ss 3144	. . . . . . . . . . . 12 ⊢ (ran 𝐸 ⊆ 𝑌 ↔ (ran 𝐸 ∩ 𝑌) = ran 𝐸)
6		incom 3329	. . . . . . . . . . . . . 14 ⊢ (ran 𝐸 ∩ 𝑌) = (𝑌 ∩ ran 𝐸)
7	6	eqeq1i 2185	. . . . . . . . . . . . 13 ⊢ ((ran 𝐸 ∩ 𝑌) = ran 𝐸 ↔ (𝑌 ∩ ran 𝐸) = ran 𝐸)
8		eqtr2 2196	. . . . . . . . . . . . . 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 4846	. . . . . 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 2184	. . . . . . 7 ⊢ (𝑋 = dom 𝐸 → (𝑋 = ∅ ↔ dom 𝐸 = ∅))
20	19	eqcoms 2180	. . . . . 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 1353   ∈ wcel 2148  ∀wral 2455  ∃wrex 2456   ∩ cin 3130   ⊆ wss 3131  ∅c0 3424  dom cdm 4628  ran crn 4629  ⟶wf 5214

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-14 2151  ax-ext 2159  ax-sep 4123  ax-pow 4176  ax-pr 4211

This theorem depends on definitions:  df-bi 117  df-3an 980  df-tru 1356  df-fal 1359  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ral 2460  df-rex 2461  df-v 2741  df-dif 3133  df-un 3135  df-in 3137  df-ss 3144  df-nul 3425  df-pw 3579  df-sn 3600  df-pr 3601  df-op 3603  df-br 4006  df-opab 4067  df-cnv 4636  df-dm 4638  df-rn 4639  df-fn 5221  df-f 5222

This theorem is referenced by: (None)