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Theorem asclpropd 15149
Description: If two structures have the same components (properties), one is an associative algebra iff the other one is. The last hypotheses on 1r can be discharged either by letting 𝑊 = V (if strong equality is known on ·𝑠) or assuming 𝐾 is a ring. (Contributed by Mario Carneiro, 5-Jul-2015.)
Hypotheses
Ref Expression
asclpropd.f 𝐹 = (Scalar‘𝐾)
asclpropd.g 𝐺 = (Scalar‘𝐿)
asclpropd.1 (𝜑 → 𝑃 = (Base‘𝐹))
asclpropd.2 (𝜑 → 𝑃 = (Base‘𝐺))
asclpropd.3 ((𝜑 ∧ (𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝑊)) → (𝑥( ·𝑠 ‘𝐾)𝑦) = (𝑥( ·𝑠 ‘𝐿)𝑦))
asclpropd.4 (𝜑 → (1r‘𝐾) = (1r‘𝐿))
asclpropd.5 (𝜑 → (1r‘𝐾) ∈ 𝑊)
Assertion
Ref Expression
asclpropd (𝜑 → (algSc‘𝐾) = (algSc‘𝐿))
Distinct variable groups:   𝑥,𝑦,𝐾   𝑥,𝐿,𝑦   𝑥,𝑃,𝑦   𝜑,𝑥,𝑦   𝑥,𝑊,𝑦
Allowed substitution hints:   𝐹(𝑥, 𝑦)   𝐺(𝑥, 𝑦)

Proof of Theorem asclpropd
Dummy variable 𝑧 is distinct from all other variables.
StepHypRef Expression
1 asclpropd.5 . . . . . 6 (𝜑 → (1r‘𝐾) ∈ 𝑊)
2 asclpropd.3 . . . . . . . 8 ((𝜑 ∧ (𝑥 ∈ 𝑃 ∧ 𝑦 ∈ 𝑊)) → (𝑥( ·𝑠 ‘𝐾)𝑦) = (𝑥( ·𝑠 ‘𝐿)𝑦))
32oveqrspc2v 6115 . . . . . . 7 ((𝜑 ∧ (𝑧 ∈ 𝑃 ∧ (1r‘𝐾) ∈ 𝑊)) → (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐾)))
43anassrs 404 . . . . . 6 (((𝜑 ∧ 𝑧 ∈ 𝑃) ∧ (1r‘𝐾) ∈ 𝑊) → (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐾)))
51, 4mpidan 427 . . . . 5 ((𝜑 ∧ 𝑧 ∈ 𝑃) → (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐾)))
6 asclpropd.4 . . . . . . 7 (𝜑 → (1r‘𝐾) = (1r‘𝐿))
76oveq2d 6104 . . . . . 6 (𝜑 → (𝑧( ·𝑠 ‘𝐿)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿)))
87adantr 276 . . . . 5 ((𝜑 ∧ 𝑧 ∈ 𝑃) → (𝑧( ·𝑠 ‘𝐿)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿)))
95, 8eqtrd 2271 . . . 4 ((𝜑 ∧ 𝑧 ∈ 𝑃) → (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾)) = (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿)))
109mpteq2dva 4223 . . 3 (𝜑 → (𝑧 ∈ 𝑃 ↦ (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾))) = (𝑧 ∈ 𝑃 ↦ (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿))))
11 asclpropd.1 . . . 4 (𝜑 → 𝑃 = (Base‘𝐹))
1211mpteq1d 4217 . . 3 (𝜑 → (𝑧 ∈ 𝑃 ↦ (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾))) = (𝑧 ∈ (Base‘𝐹) ↦ (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾))))
13 asclpropd.2 . . . 4 (𝜑 → 𝑃 = (Base‘𝐺))
1413mpteq1d 4217 . . 3 (𝜑 → (𝑧 ∈ 𝑃 ↦ (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿))) = (𝑧 ∈ (Base‘𝐺) ↦ (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿))))
1510, 12, 143eqtr3d 2279 . 2 (𝜑 → (𝑧 ∈ (Base‘𝐹) ↦ (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾))) = (𝑧 ∈ (Base‘𝐺) ↦ (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿))))
16 eqid 2238 . . 3 (algSc‘𝐾) = (algSc‘𝐾)
17 asclpropd.f . . 3 𝐹 = (Scalar‘𝐾)
18 eqid 2238 . . 3 (Base‘𝐹) = (Base‘𝐹)
19 eqid 2238 . . 3 ( ·𝑠 ‘𝐾) = ( ·𝑠 ‘𝐾)
20 eqid 2238 . . 3 (1r‘𝐾) = (1r‘𝐾)
2116, 17, 18, 19, 20asclfval 15130 . 2 (algSc‘𝐾) = (𝑧 ∈ (Base‘𝐹) ↦ (𝑧( ·𝑠 ‘𝐾)(1r‘𝐾)))
22 eqid 2238 . . 3 (algSc‘𝐿) = (algSc‘𝐿)
23 asclpropd.g . . 3 𝐺 = (Scalar‘𝐿)
24 eqid 2238 . . 3 (Base‘𝐺) = (Base‘𝐺)
25 eqid 2238 . . 3 ( ·𝑠 ‘𝐿) = ( ·𝑠 ‘𝐿)
26 eqid 2238 . . 3 (1r‘𝐿) = (1r‘𝐿)
2722, 23, 24, 25, 26asclfval 15130 . 2 (algSc‘𝐿) = (𝑧 ∈ (Base‘𝐺) ↦ (𝑧( ·𝑠 ‘𝐿)(1r‘𝐿)))
2815, 21, 273eqtr4g 2296 1 (𝜑 → (algSc‘𝐾) = (algSc‘𝐿))
Colors of variables:    wff set class
This proof depends on syntax axioms:   → wi 4   ∧ wa 104   = wceq 1402   ∈ wcel 2209   ↦ cmpt 4193  ‘cfv 5380  (class class class)co 6088  Basecbs 13410  Scalarcsca 13493   ·𝑠 cvsca 13494  1rcur 14369  algSccascl 15107
This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  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-pow 4313  ax-pr 4348  ax-un 4580  ax-cnex 8276  ax-resscn 8277  ax-1re 8279  ax-addrcl 8282
This proof depends on definitions:  df-bi 117  df-3an 1011  df-tru 1405  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-ral 2533  df-rex 2534  df-reu 2535  df-rab 2537  df-v 2823  df-sbc 3052  df-csb 3148  df-un 3224  df-in 3226  df-ss 3233  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-id 4440  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-inn 9314  df-2 9372  df-3 9373  df-4 9374  df-5 9375  df-ndx 13413  df-slot 13414  df-base 13416  df-sca 13506  df-ascl 15110
This theorem is used by: (None)
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