Theorem fneqeql2 5605

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 5604	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹 ∩ 𝐺) = 𝐴))
2		eqss 3162	. . 3 ⊢ (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
3		inss1 3347	. . . . . 6 ⊢ (𝐹 ∩ 𝐺) ⊆ 𝐹
4		dmss 4810	. . . . . 6 ⊢ ((𝐹 ∩ 𝐺) ⊆ 𝐹 → dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹)
5	3, 4	ax-mp 5	. . . . 5 ⊢ dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹
6		fndm 5297	. . . . . 6 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
7	6	adantr 274	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom 𝐹 = 𝐴)
8	5, 7	sseqtrid 3197	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom (𝐹 ∩ 𝐺) ⊆ 𝐴)
9	8	biantrurd 303	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐴 ⊆ dom (𝐹 ∩ 𝐺) ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺))))
10	2, 9	bitr4id 198	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
11	1, 10	bitrd 187	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))

Syntax hints:   → wi 4   ∧ wa 103   ↔ wb 104   = wceq 1348   ∩ cin 3120   ⊆ wss 3121  dom cdm 4611   Fn wfn 5193

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-14 2144  ax-ext 2152  ax-sep 4107  ax-pow 4160  ax-pr 4194

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-eu 2022  df-mo 2023  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-ral 2453  df-rex 2454  df-rab 2457  df-v 2732  df-sbc 2956  df-csb 3050  df-un 3125  df-in 3127  df-ss 3134  df-pw 3568  df-sn 3589  df-pr 3590  df-op 3592  df-uni 3797  df-br 3990  df-opab 4051  df-mpt 4052  df-id 4278  df-xp 4617  df-rel 4618  df-cnv 4619  df-co 4620  df-dm 4621  df-iota 5160  df-fun 5200  df-fn 5201  df-fv 5206

This theorem is referenced by: (None)