Theorem fconst5 7204

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 5918	. . . 4 ⊢ (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = ran (𝐴 × {𝐵}))
2		rnxp 6161	. . . . 5 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × {𝐵}) = {𝐵})
3	2	eqeq2d 2772	. . . 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 6537	. . . . . . 7 ⊢ (𝐹:𝐴–onto→{𝐵} ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}))
7		fof 6788	. . . . . . 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 7201	. . . . . 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 6633	. . . . 5 ⊢ (𝐹 Fn 𝐴 → Rel 𝐹)
14		snprc 4678	. . . . . 6 ⊢ (¬ 𝐵 ∈ V ↔ {𝐵} = ∅)
15		relrn0 5955	. . . . . . . . . 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 2773	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
19	18	adantr 486	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
20		xpeq2 5672	. . . . . . . . . . 11 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = (𝐴 × ∅))
21		xp0 5751	. . . . . . . . . . 11 ⊢ (𝐴 × ∅) = ∅
22	20, 21	eqtrdi 2812	. . . . . . . . . 10 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = ∅)
23	22	eqeq2d 2772	. . . . . . . . 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 2145   ≠ wne 2956  Vcvv 3451  ∅c0 4279  {csn 4584   × cxp 5649  ran crn 5652  Rel wrel 5656   Fn wfn 6526  ⟶wf 6527  –onto→wfo 6529

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 2147  ax-9 2155  ax-10 2178  ax-11 2194  ax-12 2213  ax-ext 2733  ax-sep 5249  ax-nul 5260  ax-pr 5391

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 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3078  df-rex 3088  df-rab 3414  df-v 3453  df-sbc 3740  df-csb 3848  df-dif 3902  df-un 3904  df-in 3906  df-ss 3916  df-nul 4280  df-if 4483  df-sn 4585  df-pr 4587  df-op 4591  df-uni 4868  df-br 5104  df-opab 5168  df-mpt 5187  df-id 5546  df-xp 5657  df-rel 5658  df-cnv 5659  df-co 5660  df-dm 5661  df-rn 5662  df-res 5663  df-ima 5664  df-iota 6487  df-fun 6533  df-fn 6534  df-f 6535  df-fo 6537  df-fv 6539

This theorem is used by:  imadrhmcl  21034  nvo00  31345  zar0ring  34492  esumnul  34662  esum0  34663  volsupnfl  38551