Theorem fneqeql2 5297

Description: Two functions are equal iff their equalizer contains the whole domain. (Contributed by Stefan O'Rear, 9-Mar-2015.)

Assertion

Ref	Expression
fneqeql2	⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))

Proof of Theorem fneqeql2

Step	Hyp	Ref	Expression
1		fneqeql 5296	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹 ∩ 𝐺) = 𝐴))
2		inss1 3186	. . . . . 6 ⊢ (𝐹 ∩ 𝐺) ⊆ 𝐹
3		dmss 4552	. . . . . 6 ⊢ ((𝐹 ∩ 𝐺) ⊆ 𝐹 → dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹)
4	2, 3	ax-mp 7	. . . . 5 ⊢ dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹
5		fndm 5018	. . . . . 6 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
6	5	adantr 270	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom 𝐹 = 𝐴)
7	4, 6	syl5sseq 3047	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom (𝐹 ∩ 𝐺) ⊆ 𝐴)
8	7	biantrurd 299	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐴 ⊆ dom (𝐹 ∩ 𝐺) ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺))))
9		eqss 3014	. . 3 ⊢ (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
10	8, 9	syl6rbbr 197	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
11	1, 10	bitrd 186	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))

Syntax hints:   → wi 4   ∧ wa 102   ↔ wb 103   = wceq 1284   ∩ cin 2972   ⊆ wss 2973  dom cdm 4363   Fn wfn 4917

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 662  ax-5 1376  ax-7 1377  ax-gen 1378  ax-ie1 1422  ax-ie2 1423  ax-8 1435  ax-10 1436  ax-11 1437  ax-i12 1438  ax-bndl 1439  ax-4 1440  ax-14 1445  ax-17 1459  ax-i9 1463  ax-ial 1467  ax-i5r 1468  ax-ext 2063  ax-sep 3896  ax-pow 3948  ax-pr 3964

This theorem depends on definitions:  df-bi 115  df-3an 921  df-tru 1287  df-nf 1390  df-sb 1686  df-eu 1944  df-mo 1945  df-clab 2068  df-cleq 2074  df-clel 2077  df-nfc 2208  df-ral 2353  df-rex 2354  df-rab 2357  df-v 2603  df-sbc 2816  df-csb 2909  df-un 2977  df-in 2979  df-ss 2986  df-pw 3384  df-sn 3404  df-pr 3405  df-op 3407  df-uni 3602  df-br 3786  df-opab 3840  df-mpt 3841  df-id 4048  df-xp 4369  df-rel 4370  df-cnv 4371  df-co 4372  df-dm 4373  df-iota 4887  df-fun 4924  df-fn 4925  df-fv 4930

This theorem is referenced by: (None)