Theorem fconst5 7211

Description: Two ways to express that a function is constant. (Contributed by NM, 27-Nov-2007.)

Assertion

Ref	Expression
fconst5	⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (𝐹 = (𝐴 × {𝐵}) ↔ ran 𝐹 = {𝐵}))

Proof of Theorem fconst5

Step	Hyp	Ref	Expression
1		rneq 5931	. . . 4 ⊢ (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = ran (𝐴 × {𝐵}))
2		rnxp 6173	. . . . 5 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × {𝐵}) = {𝐵})
3	2	eqeq2d 2777	. . . 4 ⊢ (𝐴 ≠ ∅ → (ran 𝐹 = ran (𝐴 × {𝐵}) ↔ ran 𝐹 = {𝐵}))
4	1, 3	imbitrid 247	. . 3 ⊢ (𝐴 ≠ ∅ → (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = {𝐵}))
5	4	adantl 487	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = {𝐵}))
6		df-fo 6549	. . . . . . 7 ⊢ (𝐹:𝐴–onto→{𝐵} ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}))
7		fof 6799	. . . . . . 7 ⊢ (𝐹:𝐴–onto→{𝐵} → 𝐹:𝐴⟶{𝐵})
8	6, 7	sylbir 238	. . . . . 6 ⊢ ((𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}) → 𝐹:𝐴⟶{𝐵})
9		fconst2g 7208	. . . . . 6 ⊢ (𝐵 ∈ V → (𝐹:𝐴⟶{𝐵} ↔ 𝐹 = (𝐴 × {𝐵})))
10	8, 9	imbitrid 247	. . . . 5 ⊢ (𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}) → 𝐹 = (𝐴 × {𝐵})))
11	10	expd 421	. . . 4 ⊢ (𝐵 ∈ V → (𝐹 Fn 𝐴 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
12	11	adantrd 497	. . 3 ⊢ (𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
13		fnrel 6644	. . . . 5 ⊢ (𝐹 Fn 𝐴 → Rel 𝐹)
14		snprc 4688	. . . . . 6 ⊢ (¬ 𝐵 ∈ V ↔ {𝐵} = ∅)
15		relrn0 5968	. . . . . . . . . 10 ⊢ (Rel 𝐹 → (𝐹 = ∅ ↔ ran 𝐹 = ∅))
16	15	biimprd 251	. . . . . . . . 9 ⊢ (Rel 𝐹 → (ran 𝐹 = ∅ → 𝐹 = ∅))
17	16	adantl 487	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = ∅ → 𝐹 = ∅))
18		eqeq2 2778	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
19	18	adantr 486	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
20		xpeq2 5687	. . . . . . . . . . 11 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = (𝐴 × ∅))
21		xp0 5766	. . . . . . . . . . 11 ⊢ (𝐴 × ∅) = ∅
22	20, 21	eqtrdi 2817	. . . . . . . . . 10 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = ∅)
23	22	eqeq2d 2777	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (𝐹 = (𝐴 × {𝐵}) ↔ 𝐹 = ∅))
24	23	adantr 486	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (𝐹 = (𝐴 × {𝐵}) ↔ 𝐹 = ∅))
25	17, 19, 24	3imtr4d 297	. . . . . . 7 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵})))
26	25	ex 418	. . . . . 6 ⊢ ({𝐵} = ∅ → (Rel 𝐹 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
27	14, 26	sylbi 220	. . . . 5 ⊢ (¬ 𝐵 ∈ V → (Rel 𝐹 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
28	13, 27	syl5 35	. . . 4 ⊢ (¬ 𝐵 ∈ V → (𝐹 Fn 𝐴 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
29	28	adantrd 497	. . 3 ⊢ (¬ 𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
30	12, 29	pm2.61i 184	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵})))
31	5, 30	impbid 215	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (𝐹 = (𝐴 × {𝐵}) ↔ ran 𝐹 = {𝐵}))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209   ∧ wa 401   = wceq 1570   ∈ wcel 2146   ≠ wne 2961  Vcvv 3458  ∅c0 4289  {csn 4594   × cxp 5664  ran crn 5667  Rel wrel 5671   Fn wfn 6538  ⟶wf 6539  –onto→wfo 6541

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 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-10 2179  ax-11 2195  ax-12 2216  ax-ext 2738  ax-sep 5262  ax-nul 5274  ax-pr 5409

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-nf 1817  df-sb 2100  df-mo 2570  df-eu 2600  df-clab 2745  df-cleq 2758  df-clel 2841  df-nfc 2915  df-ne 2962  df-ral 3083  df-rex 3093  df-rab 3420  df-v 3460  df-sbc 3748  df-csb 3857  df-dif 3911  df-un 3913  df-in 3915  df-ss 3925  df-nul 4290  df-if 4493  df-sn 4595  df-pr 4597  df-op 4601  df-uni 4878  df-br 5115  df-opab 5179  df-mpt 5198  df-id 5561  df-xp 5672  df-rel 5673  df-cnv 5674  df-co 5675  df-dm 5676  df-rn 5677  df-res 5678  df-ima 5679  df-iota 6499  df-fun 6545  df-fn 6546  df-f 6547  df-fo 6549  df-fv 6551

This theorem is used by:  imadrhmcl  20937  nvo00  31150  zar0ring  34299  esumnul  34469  esum0  34470  volsupnfl  38357