Theorem recmulnqg 7765

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 6095	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ·Q 𝑦) = (𝐴 ·Q 𝑦))
2	1	eqeq1d 2247	. . . 4 ⊢ (𝑥 = 𝐴 → ((𝑥 ·Q 𝑦) = 1Q ↔ (𝐴 ·Q 𝑦) = 1Q))
3	2	anbi2d 468	. . 3 ⊢ (𝑥 = 𝐴 → ((𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑦 ∈ Q ∧ (𝐴 ·Q 𝑦) = 1Q)))
4		eleq1 2301	. . . 4 ⊢ (𝑦 = 𝐵 → (𝑦 ∈ Q ↔ 𝐵 ∈ Q))
5		oveq2 6096	. . . . 5 ⊢ (𝑦 = 𝐵 → (𝐴 ·Q 𝑦) = (𝐴 ·Q 𝐵))
6	5	eqeq1d 2247	. . . 4 ⊢ (𝑦 = 𝐵 → ((𝐴 ·Q 𝑦) = 1Q ↔ (𝐴 ·Q 𝐵) = 1Q))
7	4, 6	anbi12d 477	. . 3 ⊢ (𝑦 = 𝐵 → ((𝑦 ∈ Q ∧ (𝐴 ·Q 𝑦) = 1Q) ↔ (𝐵 ∈ Q ∧ (𝐴 ·Q 𝐵) = 1Q)))
8		recexnq 7764	. . . 4 ⊢ (𝑥 ∈ Q → ∃𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
9		1nq 7740	. . . . 5 ⊢ 1Q ∈ Q
10		mulcomnqg 7757	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q) → (𝑧 ·Q 𝑤) = (𝑤 ·Q 𝑧))
11		mulassnqg 7758	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q ∧ 𝑣 ∈ Q) → ((𝑧 ·Q 𝑤) ·Q 𝑣) = (𝑧 ·Q (𝑤 ·Q 𝑣)))
12		mulidnq 7763	. . . . 5 ⊢ (𝑧 ∈ Q → (𝑧 ·Q 1Q) = 𝑧)
13	9, 10, 11, 12	caovimo 6286	. . . 4 ⊢ (𝑥 ∈ Q → ∃*𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
14		eu5 2134	. . . 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 421	. . 3 ⊢ (𝑥 ∈ Q → ∃!𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
16		df-rq 7726	. . . 4 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)}
17		3anass 1013	. . . . 5 ⊢ ((𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)))
18	17	opabbii 4199	. . . 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 2259	. . 3 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))}
20	3, 7, 15, 19	fvopab3g 5781	. 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))

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105   ∧ w3a 1009   = wceq 1402  ∃wex 1545  ∃!weu 2086  ∃*wmo 2087   ∈ wcel 2209  {copab 4192  ‘cfv 5380  (class class class)co 6088  Qcnq 7654  1_Qc1q 7655   ·_Q cmq 7657  *_Qcrq 7658

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-14 2212  ax-ext 2220  ax-coll 4248  ax-sep 4251  ax-nul 4261  ax-pow 4313  ax-pr 4348  ax-un 4580  ax-setind 4686  ax-iinf 4737

This proof depends on definitions:  df-bi 117  df-dc 847  df-3or 1010  df-3an 1011  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-eu 2089  df-mo 2090  df-clab 2225  df-cleq 2231  df-clel 2234  df-nfc 2381  df-ne 2421  df-ral 2533  df-rex 2534  df-reu 2535  df-rab 2537  df-v 2823  df-sbc 3052  df-csb 3148  df-dif 3222  df-un 3224  df-in 3226  df-ss 3233  df-nul 3521  df-pw 3690  df-sn 3715  df-pr 3716  df-op 3718  df-uni 3936  df-int 3971  df-iun 4014  df-br 4132  df-opab 4194  df-mpt 4195  df-tr 4232  df-id 4440  df-iord 4513  df-on 4515  df-suc 4518  df-iom 4740  df-xp 4782  df-rel 4783  df-cnv 4784  df-co 4785  df-dm 4786  df-rn 4787  df-res 4788  df-ima 4789  df-iota 5340  df-fun 5382  df-fn 5383  df-f 5384  df-f1 5385  df-fo 5386  df-f1o 5387  df-fv 5388  df-ov 6091  df-oprab 6092  df-mpo 6093  df-1st 6378  df-2nd 6379  df-recs 6580  df-irdg 6645  df-1o 6691  df-oadd 6695  df-omul 6696  df-er 6811  df-ec 6813  df-qs 6817  df-ni 7678  df-mi 7680  df-mpq 7719  df-enq 7721  df-nqqs 7722  df-mqqs 7724  df-1nqqs 7725  df-rq 7726

This theorem is used by:  recclnq  7766  recidnq  7767  recrecnq  7768  recexprlem1ssl  8007  recexprlem1ssu  8008