Theorem fneqeql2 5671

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 5670	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹 ∩ 𝐺) = 𝐴))
2		eqss 3198	. . 3 ⊢ (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
3		inss1 3383	. . . . . 6 ⊢ (𝐹 ∩ 𝐺) ⊆ 𝐹
4		dmss 4865	. . . . . 6 ⊢ ((𝐹 ∩ 𝐺) ⊆ 𝐹 → dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹)
5	3, 4	ax-mp 5	. . . . 5 ⊢ dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹
6		fndm 5357	. . . . . 6 ⊢ (𝐹 Fn 𝐴 → dom 𝐹 = 𝐴)
7	6	adantr 276	. . . . 5 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom 𝐹 = 𝐴)
8	5, 7	sseqtrid 3233	. . . 4 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → dom (𝐹 ∩ 𝐺) ⊆ 𝐴)
9	8	biantrurd 305	. . 3 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐴 ⊆ dom (𝐹 ∩ 𝐺) ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺))))
10	2, 9	bitr4id 199	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
11	1, 10	bitrd 188	1 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   = wceq 1364   ∩ cin 3156   ⊆ wss 3157  dom cdm 4663   Fn wfn 5253

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 710  ax-5 1461  ax-7 1462  ax-gen 1463  ax-ie1 1507  ax-ie2 1508  ax-8 1518  ax-10 1519  ax-11 1520  ax-i12 1521  ax-bndl 1523  ax-4 1524  ax-17 1540  ax-i9 1544  ax-ial 1548  ax-i5r 1549  ax-14 2170  ax-ext 2178  ax-sep 4151  ax-pow 4207  ax-pr 4242

This theorem depends on definitions:  df-bi 117  df-3an 982  df-tru 1367  df-nf 1475  df-sb 1777  df-eu 2048  df-mo 2049  df-clab 2183  df-cleq 2189  df-clel 2192  df-nfc 2328  df-ral 2480  df-rex 2481  df-rab 2484  df-v 2765  df-sbc 2990  df-csb 3085  df-un 3161  df-in 3163  df-ss 3170  df-pw 3607  df-sn 3628  df-pr 3629  df-op 3631  df-uni 3840  df-br 4034  df-opab 4095  df-mpt 4096  df-id 4328  df-xp 4669  df-rel 4670  df-cnv 4671  df-co 4672  df-dm 4673  df-iota 5219  df-fun 5260  df-fn 5261  df-fv 5266

This theorem is referenced by: (None)