Theorem fneqeql2 5569

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 5568	. 2 ⊢ ((𝐹 Fn 𝐴 ∧ 𝐺 Fn 𝐴) → (𝐹 = 𝐺 ↔ dom (𝐹 ∩ 𝐺) = 𝐴))
2		eqss 3139	. . 3 ⊢ (dom (𝐹 ∩ 𝐺) = 𝐴 ↔ (dom (𝐹 ∩ 𝐺) ⊆ 𝐴 ∧ 𝐴 ⊆ dom (𝐹 ∩ 𝐺)))
3		inss1 3323	. . . . . 6 ⊢ (𝐹 ∩ 𝐺) ⊆ 𝐹
4		dmss 4778	. . . . . 6 ⊢ ((𝐹 ∩ 𝐺) ⊆ 𝐹 → dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹)
5	3, 4	ax-mp 5	. . . . 5 ⊢ dom (𝐹 ∩ 𝐺) ⊆ dom 𝐹
6		fndm 5262	. . . . . 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 3174	. . . 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 1332   ∩ cin 3097   ⊆ wss 3098  dom cdm 4579   Fn wfn 5158

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 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1481  ax-10 1482  ax-11 1483  ax-i12 1484  ax-bndl 1486  ax-4 1487  ax-17 1503  ax-i9 1507  ax-ial 1511  ax-i5r 1512  ax-14 2128  ax-ext 2136  ax-sep 4078  ax-pow 4130  ax-pr 4164

This theorem depends on definitions:  df-bi 116  df-3an 965  df-tru 1335  df-nf 1438  df-sb 1740  df-eu 2006  df-mo 2007  df-clab 2141  df-cleq 2147  df-clel 2150  df-nfc 2285  df-ral 2437  df-rex 2438  df-rab 2441  df-v 2711  df-sbc 2934  df-csb 3028  df-un 3102  df-in 3104  df-ss 3111  df-pw 3541  df-sn 3562  df-pr 3563  df-op 3565  df-uni 3769  df-br 3962  df-opab 4022  df-mpt 4023  df-id 4248  df-xp 4585  df-rel 4586  df-cnv 4587  df-co 4588  df-dm 4589  df-iota 5128  df-fun 5165  df-fn 5166  df-fv 5171

This theorem is referenced by: (None)