Theorem fconst5 7186

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 5910	. . . 4 ⊢ (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = ran (𝐴 × {𝐵}))
2		rnxp 6152	. . . . 5 ⊢ (𝐴 ≠ ∅ → ran (𝐴 × {𝐵}) = {𝐵})
3	2	eqeq2d 2772	. . . 4 ⊢ (𝐴 ≠ ∅ → (ran 𝐹 = ran (𝐴 × {𝐵}) ↔ ran 𝐹 = {𝐵}))
4	1, 3	imbitrid 246	. . 3 ⊢ (𝐴 ≠ ∅ → (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = {𝐵}))
5	4	adantl 485	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (𝐹 = (𝐴 × {𝐵}) → ran 𝐹 = {𝐵}))
6		df-fo 6523	. . . . . . 7 ⊢ (𝐹:𝐴–onto→{𝐵} ↔ (𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}))
7		fof 6774	. . . . . . 7 ⊢ (𝐹:𝐴–onto→{𝐵} → 𝐹:𝐴⟶{𝐵})
8	6, 7	sylbir 237	. . . . . 6 ⊢ ((𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}) → 𝐹:𝐴⟶{𝐵})
9		fconst2g 7183	. . . . . 6 ⊢ (𝐵 ∈ V → (𝐹:𝐴⟶{𝐵} ↔ 𝐹 = (𝐴 × {𝐵})))
10	8, 9	imbitrid 246	. . . . 5 ⊢ (𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ ran 𝐹 = {𝐵}) → 𝐹 = (𝐴 × {𝐵})))
11	10	expd 419	. . . 4 ⊢ (𝐵 ∈ V → (𝐹 Fn 𝐴 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
12	11	adantrd 495	. . 3 ⊢ (𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
13		fnrel 6619	. . . . 5 ⊢ (𝐹 Fn 𝐴 → Rel 𝐹)
14		snprc 4675	. . . . . 6 ⊢ (¬ 𝐵 ∈ V ↔ {𝐵} = ∅)
15		relrn0 5947	. . . . . . . . . 10 ⊢ (Rel 𝐹 → (𝐹 = ∅ ↔ ran 𝐹 = ∅))
16	15	biimprd 250	. . . . . . . . 9 ⊢ (Rel 𝐹 → (ran 𝐹 = ∅ → 𝐹 = ∅))
17	16	adantl 485	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = ∅ → 𝐹 = ∅))
18		eqeq2 2773	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
19	18	adantr 484	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} ↔ ran 𝐹 = ∅))
20		xpeq2 5666	. . . . . . . . . . 11 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = (𝐴 × ∅))
21		xp0 5745	. . . . . . . . . . 11 ⊢ (𝐴 × ∅) = ∅
22	20, 21	eqtrdi 2812	. . . . . . . . . 10 ⊢ ({𝐵} = ∅ → (𝐴 × {𝐵}) = ∅)
23	22	eqeq2d 2772	. . . . . . . . 9 ⊢ ({𝐵} = ∅ → (𝐹 = (𝐴 × {𝐵}) ↔ 𝐹 = ∅))
24	23	adantr 484	. . . . . . . 8 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (𝐹 = (𝐴 × {𝐵}) ↔ 𝐹 = ∅))
25	17, 19, 24	3imtr4d 296	. . . . . . 7 ⊢ (({𝐵} = ∅ ∧ Rel 𝐹) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵})))
26	25	ex 416	. . . . . 6 ⊢ ({𝐵} = ∅ → (Rel 𝐹 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
27	14, 26	sylbi 219	. . . . 5 ⊢ (¬ 𝐵 ∈ V → (Rel 𝐹 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
28	13, 27	syl5 34	. . . 4 ⊢ (¬ 𝐵 ∈ V → (𝐹 Fn 𝐴 → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
29	28	adantrd 495	. . 3 ⊢ (¬ 𝐵 ∈ V → ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵}))))
30	12, 29	pm2.61i 183	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (ran 𝐹 = {𝐵} → 𝐹 = (𝐴 × {𝐵})))
31	5, 30	impbid 214	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐴 ≠ ∅) → (𝐹 = (𝐴 × {𝐵}) ↔ ran 𝐹 = {𝐵}))

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 208   ∧ wa 399   = wceq 1559   ∈ wcel 2141   ≠ wne 2956  Vcvv 3453  ∅c0 4285  {csn 4581   × cxp 5643  ran crn 5646  Rel wrel 5650   Fn wfn 6512  ⟶wf 6513  –onto→wfo 6515

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-11 2190  ax-12 2211  ax-ext 2733  ax-sep 5245  ax-nul 5255  ax-pr 5389

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-3an 1099  df-tru 1562  df-fal 1572  df-ex 1799  df-nf 1803  df-sb 2090  df-mo 2565  df-eu 2595  df-clab 2740  df-cleq 2753  df-clel 2836  df-nfc 2910  df-ne 2957  df-ral 3076  df-rex 3086  df-rab 3414  df-v 3455  df-sbc 3745  df-csb 3853  df-dif 3907  df-un 3909  df-in 3911  df-ss 3921  df-nul 4286  df-if 4480  df-sn 4582  df-pr 4584  df-op 4588  df-uni 4865  df-br 5100  df-opab 5162  df-mpt 5181  df-id 5540  df-xp 5651  df-rel 5652  df-cnv 5653  df-co 5654  df-dm 5655  df-rn 5656  df-res 5657  df-ima 5658  df-iota 6473  df-fun 6519  df-fn 6520  df-f 6521  df-fo 6523  df-fv 6525

This theorem is referenced by:  imadrhmcl  20826  nvo00  30910  zar0ring  34136  esumnul  34306  esum0  34307  volsupnfl  38128