Theorem recmulnqg 7392

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 5884	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝑥 ·Q 𝑦) = (𝐴 ·Q 𝑦))
2	1	eqeq1d 2186	. . . 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 2240	. . . 4 ⊢ (𝑦 = 𝐵 → (𝑦 ∈ Q ↔ 𝐵 ∈ Q))
5		oveq2 5885	. . . . 5 ⊢ (𝑦 = 𝐵 → (𝐴 ·Q 𝑦) = (𝐴 ·Q 𝐵))
6	5	eqeq1d 2186	. . . 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 7391	. . . 4 ⊢ (𝑥 ∈ Q → ∃𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
9		1nq 7367	. . . . 5 ⊢ 1Q ∈ Q
10		mulcomnqg 7384	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q) → (𝑧 ·Q 𝑤) = (𝑤 ·Q 𝑧))
11		mulassnqg 7385	. . . . 5 ⊢ ((𝑧 ∈ Q ∧ 𝑤 ∈ Q ∧ 𝑣 ∈ Q) → ((𝑧 ·Q 𝑤) ·Q 𝑣) = (𝑧 ·Q (𝑤 ·Q 𝑣)))
12		mulidnq 7390	. . . . 5 ⊢ (𝑧 ∈ Q → (𝑧 ·Q 1Q) = 𝑧)
13	9, 10, 11, 12	caovimo 6070	. . . 4 ⊢ (𝑥 ∈ Q → ∃*𝑦(𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))
14		eu5 2073	. . . 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 7353	. . . 4 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)}
17		3anass 982	. . . . 5 ⊢ ((𝑥 ∈ Q ∧ 𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q) ↔ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q)))
18	17	opabbii 4072	. . . 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 2198	. . 3 ⊢ *Q = {⟨𝑥, 𝑦⟩ ∣ (𝑥 ∈ Q ∧ (𝑦 ∈ Q ∧ (𝑥 ·Q 𝑦) = 1Q))}
20	3, 7, 15, 19	fvopab3g 5591	. 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 978   = wceq 1353  ∃wex 1492  ∃!weu 2026  ∃*wmo 2027   ∈ wcel 2148  {copab 4065  ‘cfv 5218  (class class class)co 5877  Qcnq 7281  1_Qc1q 7282   ·_Q cmq 7284  *_Qcrq 7285

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 614  ax-in2 615  ax-io 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-13 2150  ax-14 2151  ax-ext 2159  ax-coll 4120  ax-sep 4123  ax-nul 4131  ax-pow 4176  ax-pr 4211  ax-un 4435  ax-setind 4538  ax-iinf 4589

This theorem depends on definitions:  df-bi 117  df-dc 835  df-3or 979  df-3an 980  df-tru 1356  df-fal 1359  df-nf 1461  df-sb 1763  df-eu 2029  df-mo 2030  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-ne 2348  df-ral 2460  df-rex 2461  df-reu 2462  df-rab 2464  df-v 2741  df-sbc 2965  df-csb 3060  df-dif 3133  df-un 3135  df-in 3137  df-ss 3144  df-nul 3425  df-pw 3579  df-sn 3600  df-pr 3601  df-op 3603  df-uni 3812  df-int 3847  df-iun 3890  df-br 4006  df-opab 4067  df-mpt 4068  df-tr 4104  df-id 4295  df-iord 4368  df-on 4370  df-suc 4373  df-iom 4592  df-xp 4634  df-rel 4635  df-cnv 4636  df-co 4637  df-dm 4638  df-rn 4639  df-res 4640  df-ima 4641  df-iota 5180  df-fun 5220  df-fn 5221  df-f 5222  df-f1 5223  df-fo 5224  df-f1o 5225  df-fv 5226  df-ov 5880  df-oprab 5881  df-mpo 5882  df-1st 6143  df-2nd 6144  df-recs 6308  df-irdg 6373  df-1o 6419  df-oadd 6423  df-omul 6424  df-er 6537  df-ec 6539  df-qs 6543  df-ni 7305  df-mi 7307  df-mpq 7346  df-enq 7348  df-nqqs 7349  df-mqqs 7351  df-1nqqs 7352  df-rq 7353

This theorem is referenced by:  recclnq  7393  recidnq  7394  recrecnq  7395  recexprlem1ssl  7634  recexprlem1ssu  7635