Theorem recmulnqg 7616

Description: Relationship between reciprocal and multiplication on positive fractions. (Contributed by Jim Kingdon, 19-Sep-2019.)

Assertion

Ref	Expression
recmulnqg	⊢ ((𝐴 ∈ Q ∧ 𝐵 ∈ Q) → ((*Q‘𝐴) = 𝐵 ↔ (𝐴 ·Q 𝐵) = 1Q))

Proof of Theorem recmulnqg

Dummy variables 𝑥 𝑦 𝑧 𝑤 𝑣 are mutually distinct and distinct from all other variables.

Step	Hyp	Ref	Expression
1		oveq1 6030	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ·Q 𝑦) = (𝐴 ·Q 𝑦))
2	1	eqeq1d 2239	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑥 ·Q 𝑦) = 1Q ↔ (𝐴 ·Q 𝑦) = 1Q))
3	2	anbi2d 464	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑦 ∈ Q ∧ (𝐴 ·Q 𝑦) = 1Q)))
4		eleq1 2293	. . . 4 ⊢ (𝑦 = 𝐵 → (𝑦 ∈ Q ↔ 𝐵 ∈ Q))
5		oveq2 6031	. . . . 5 ⊢ (𝑦 = 𝐵 → (𝐴 ·Q 𝑦) = (𝐴 ·Q 𝐵))
6	5	eqeq1d 2239	. . . 4 ⊢ (𝑦 = 𝐵 → ((𝐴 ·Q 𝑦) = 1Q ↔ (𝐴 ·Q 𝐵) = 1Q))
7	4, 6	anbi12d 473	. . 3 ⊢ (𝑦 = 𝐵 → ((𝑦 ∈ Q ∧ (𝐴 ·Q 𝑦) = 1Q) ↔ (𝐵 ∈ Q ∧ (𝐴 ·Q 𝐵) = 1Q)))
8		recexnq 7615	. . . 4 ⊢ (𝑥 ∈ Q → ∃𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
9		1nq 7591	. . . . 5 ⊢ 1Q ∈ Q
10		mulcomnqg 7608	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q) → (𝑧 ·Q 𝑤) = (𝑤 ·Q 𝑧))
11		mulassnqg 7609	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q ∧ 𝑣 ∈ Q) → ((𝑧 ·Q 𝑤) ·Q 𝑣) = (𝑧 ·Q (𝑤 ·Q 𝑣)))
12		mulidnq 7614	. . . . 5 ⊢ (𝑧 ∈ Q → (𝑧 ·Q 1Q) = 𝑧)
13	9, 10, 11, 12	caovimo 6221	. . . 4 ⊢ (𝑥 ∈ Q → ∃*𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
14		eu5 2126	. . . 4 ⊢ (∃!𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (∃𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ∧ ∃*𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)))
15	8, 13, 14	sylanbrc 417	. . 3 ⊢ (𝑥 ∈ Q → ∃!𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
16		df-rq 7577	. . . 4 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)}
17		3anass 1008	. . . . 5 ⊢ ((𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)))
18	17	opabbii 4157	. . . 4 ⊢ {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)} = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))}
19	16, 18	eqtri 2251	. . 3 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))}
20	3, 7, 15, 19	fvopab3g 5722	. 2 ⊢ ((𝐴 ∈ Q ∧ 𝐵 ∈ Q) → ((*Q‘𝐴) = 𝐵 ↔ (𝐵 ∈ Q ∧ (𝐴 ·Q 𝐵) = 1Q)))
21		ibar 301	. . 3 ⊢ (𝐵 ∈ Q → ((𝐴 ·Q 𝐵) = 1Q ↔ (𝐵 ∈ Q ∧ (𝐴 ·Q 𝐵) = 1Q)))
22	21	adantl 277	. 2 ⊢ ((𝐴 ∈ Q ∧ 𝐵 ∈ Q) → ((𝐴 ·Q 𝐵) = 1Q ↔ (𝐵 ∈ Q ∧ (𝐴 ·Q 𝐵) = 1Q)))
23	20, 22	bitr4d 191	1 ⊢ ((𝐴 ∈ Q ∧ 𝐵 ∈ Q) → ((*Q‘𝐴) = 𝐵 ↔ (𝐴 ·Q 𝐵) = 1Q))

Syntax hints:   → wi 4   ∧ wa 104   ↔ wb 105   ∧ w3a 1004   = wceq 1397  ∃wex 1540  ∃!weu 2078  ∃*wmo 2079   ∈ wcel 2201  {copab 4150  ‘cfv 5328  (class class class)co 6023  Qcnq 7505  1_Qc1q 7506   ·_Q cmq 7508  *_Qcrq 7509

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-in1 619  ax-in2 620  ax-io 716  ax-5 1495  ax-7 1496  ax-gen 1497  ax-ie1 1541  ax-ie2 1542  ax-8 1552  ax-10 1553  ax-11 1554  ax-i12 1555  ax-bndl 1557  ax-4 1558  ax-17 1574  ax-i9 1578  ax-ial 1582  ax-i5r 1583  ax-13 2203  ax-14 2204  ax-ext 2212  ax-coll 4205  ax-sep 4208  ax-nul 4216  ax-pow 4266  ax-pr 4301  ax-un 4532  ax-setind 4637  ax-iinf 4688

This theorem depends on definitions:  df-bi 117  df-dc 842  df-3or 1005  df-3an 1006  df-tru 1400  df-fal 1403  df-nf 1509  df-sb 1810  df-eu 2081  df-mo 2082  df-clab 2217  df-cleq 2223  df-clel 2226  df-nfc 2362  df-ne 2402  df-ral 2514  df-rex 2515  df-reu 2516  df-rab 2518  df-v 2803  df-sbc 3031  df-csb 3127  df-dif 3201  df-un 3203  df-in 3205  df-ss 3212  df-nul 3494  df-pw 3655  df-sn 3676  df-pr 3677  df-op 3679  df-uni 3895  df-int 3930  df-iun 3973  df-br 4090  df-opab 4152  df-mpt 4153  df-tr 4189  df-id 4392  df-iord 4465  df-on 4467  df-suc 4470  df-iom 4691  df-xp 4733  df-rel 4734  df-cnv 4735  df-co 4736  df-dm 4737  df-rn 4738  df-res 4739  df-ima 4740  df-iota 5288  df-fun 5330  df-fn 5331  df-f 5332  df-f1 5333  df-fo 5334  df-f1o 5335  df-fv 5336  df-ov 6026  df-oprab 6027  df-mpo 6028  df-1st 6308  df-2nd 6309  df-recs 6476  df-irdg 6541  df-1o 6587  df-oadd 6591  df-omul 6592  df-er 6707  df-ec 6709  df-qs 6713  df-ni 7529  df-mi 7531  df-mpq 7570  df-enq 7572  df-nqqs 7573  df-mqqs 7575  df-1nqqs 7576  df-rq 7577

This theorem is referenced by:  recclnq  7617  recidnq  7618  recrecnq  7619  recexprlem1ssl  7858  recexprlem1ssu  7859