Theorem recmulnqg 7451

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 5925	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ·Q 𝑦) = (𝐴 ·Q 𝑦))
2	1	eqeq1d 2202	. . . 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 2256	. . . 4 ⊢ (𝑦 = 𝐵 → (𝑦 ∈ Q ↔ 𝐵 ∈ Q))
5		oveq2 5926	. . . . 5 ⊢ (𝑦 = 𝐵 → (𝐴 ·Q 𝑦) = (𝐴 ·Q 𝐵))
6	5	eqeq1d 2202	. . . 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 7450	. . . 4 ⊢ (𝑥 ∈ Q → ∃𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
9		1nq 7426	. . . . 5 ⊢ 1Q ∈ Q
10		mulcomnqg 7443	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q) → (𝑧 ·Q 𝑤) = (𝑤 ·Q 𝑧))
11		mulassnqg 7444	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q ∧ 𝑣 ∈ Q) → ((𝑧 ·Q 𝑤) ·Q 𝑣) = (𝑧 ·Q (𝑤 ·Q 𝑣)))
12		mulidnq 7449	. . . . 5 ⊢ (𝑧 ∈ Q → (𝑧 ·Q 1Q) = 𝑧)
13	9, 10, 11, 12	caovimo 6112	. . . 4 ⊢ (𝑥 ∈ Q → ∃*𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
14		eu5 2089	. . . 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 7412	. . . 4 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)}
17		3anass 984	. . . . 5 ⊢ ((𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)))
18	17	opabbii 4096	. . . 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 2214	. . 3 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))}
20	3, 7, 15, 19	fvopab3g 5630	. 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 980   = wceq 1364  ∃wex 1503  ∃!weu 2042  ∃*wmo 2043   ∈ wcel 2164  {copab 4089  ‘cfv 5254  (class class class)co 5918  Qcnq 7340  1_Qc1q 7341   ·_Q cmq 7343  *_Qcrq 7344

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 615  ax-in2 616  ax-io 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-bndl 1520  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-13 2166  ax-14 2167  ax-ext 2175  ax-coll 4144  ax-sep 4147  ax-nul 4155  ax-pow 4203  ax-pr 4238  ax-un 4464  ax-setind 4569  ax-iinf 4620

This theorem depends on definitions:  df-bi 117  df-dc 836  df-3or 981  df-3an 982  df-tru 1367  df-fal 1370  df-nf 1472  df-sb 1774  df-eu 2045  df-mo 2046  df-clab 2180  df-cleq 2186  df-clel 2189  df-nfc 2325  df-ne 2365  df-ral 2477  df-rex 2478  df-reu 2479  df-rab 2481  df-v 2762  df-sbc 2986  df-csb 3081  df-dif 3155  df-un 3157  df-in 3159  df-ss 3166  df-nul 3447  df-pw 3603  df-sn 3624  df-pr 3625  df-op 3627  df-uni 3836  df-int 3871  df-iun 3914  df-br 4030  df-opab 4091  df-mpt 4092  df-tr 4128  df-id 4324  df-iord 4397  df-on 4399  df-suc 4402  df-iom 4623  df-xp 4665  df-rel 4666  df-cnv 4667  df-co 4668  df-dm 4669  df-rn 4670  df-res 4671  df-ima 4672  df-iota 5215  df-fun 5256  df-fn 5257  df-f 5258  df-f1 5259  df-fo 5260  df-f1o 5261  df-fv 5262  df-ov 5921  df-oprab 5922  df-mpo 5923  df-1st 6193  df-2nd 6194  df-recs 6358  df-irdg 6423  df-1o 6469  df-oadd 6473  df-omul 6474  df-er 6587  df-ec 6589  df-qs 6593  df-ni 7364  df-mi 7366  df-mpq 7405  df-enq 7407  df-nqqs 7408  df-mqqs 7410  df-1nqqs 7411  df-rq 7412

This theorem is referenced by:  recclnq  7452  recidnq  7453  recrecnq  7454  recexprlem1ssl  7693  recexprlem1ssu  7694