Theorem fconst5 7235

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 5925	. . . 4 ⊢ (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = ran (𝐴 × {𝐵}))
2		rnxp 6174	. . . . 5 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × {𝐵}) = {𝐵})
3	2	eqeq2d 2774	. . . 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 6564	. . . . . . 7 ⊢ (𝐹:𝐴⟶onto{𝐵} ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}))
7		fof 6818	. . . . . . 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 7232	. . . . . 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 6663	. . . . 5 ⊢ (𝐹 Fn 𝐴 → Rel 𝐹)
14		snprc 4678	. . . . . 6 ⊢ (¬ 𝐵 ∈ V ↔ {𝐵} = ∅)
15		relrn0 5962	. . . . . . . . . 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 2775	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
19	18	adantr 486	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
20		xpeq2 5676	. . . . . . . . . . 11 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = (𝐴 × ∅))
21		xp0 5756	. . . . . . . . . . 11 ⊢ (𝐴 × ∅) = ∅
22	20, 21	eqtrdi 2814	. . . . . . . . . 10 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = ∅)
23	22	eqeq2d 2774	. . . . . . . . 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 2147   ≠ wne 2959  Vcvv 3454  ∅c0 4281  {csn 4588   × cxp 5653  ran crn 5656  Rel wrel 5660   Fn wfn 6553  ⟶wf 6554  ⟶ontowfo 6556

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-10 2180  ax-11 2197  ax-12 2216  ax-ext 2735  ax-sep 5252  ax-nul 5263  ax-pr 5395

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 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-nfc 2913  df-ne 2960  df-ral 3081  df-rex 3091  df-rab 3417  df-v 3456  df-sbc 3743  df-csb 3852  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-uni 4871  df-br 5107  df-opab 5171  df-mpt 5190  df-id 5550  df-xp 5661  df-rel 5662  df-cnv 5663  df-co 5664  df-dm 5665  df-rn 5666  df-res 5667  df-ima 5668  df-iota 6512  df-fun 6560  df-fn 6561  df-f 6562  df-fo 6564  df-fv 6566

This theorem is used by:  imadrhmcl  21084  nvo00  31393  zar0ring  34580  esumnul  34750  esum0  34751  volsupnfl  38775