Theorem fneqeql2 5689

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 5688	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹 ∩ 𝐺) = 𝐴))
2		eqss 3208	. . 3 ⊢ (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
3		inss1 3393	. . . . . 6 ⊢ (𝐹 ∩ 𝐺) ⊆ 𝐹
4		dmss 4877	. . . . . 6 ⊢ ((𝐹 ∩ 𝐺) ⊆ 𝐹 → dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹)
5	3, 4	ax-mp 5	. . . . 5 ⊢ dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹
6		fndm 5373	. . . . . 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 3243	. . . 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 1373   ∩ cin 3165   ⊆ wss 3166  dom cdm 4675   Fn wfn 5266

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 711  ax-5 1470  ax-7 1471  ax-gen 1472  ax-ie1 1516  ax-ie2 1517  ax-8 1527  ax-10 1528  ax-11 1529  ax-i12 1530  ax-bndl 1532  ax-4 1533  ax-17 1549  ax-i9 1553  ax-ial 1557  ax-i5r 1558  ax-14 2179  ax-ext 2187  ax-sep 4162  ax-pow 4218  ax-pr 4253

This theorem depends on definitions:  df-bi 117  df-3an 983  df-tru 1376  df-nf 1484  df-sb 1786  df-eu 2057  df-mo 2058  df-clab 2192  df-cleq 2198  df-clel 2201  df-nfc 2337  df-ral 2489  df-rex 2490  df-rab 2493  df-v 2774  df-sbc 2999  df-csb 3094  df-un 3170  df-in 3172  df-ss 3179  df-pw 3618  df-sn 3639  df-pr 3640  df-op 3642  df-uni 3851  df-br 4045  df-opab 4106  df-mpt 4107  df-id 4340  df-xp 4681  df-rel 4682  df-cnv 4683  df-co 4684  df-dm 4685  df-iota 5232  df-fun 5273  df-fn 5274  df-fv 5279

This theorem is referenced by: (None)