Metamath Proof Explorer


Theorem asclpropd

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)

Ref Expression
Hypotheses asclpropd.f 𝐹 = ( Scalar ‘ 𝐾 )
asclpropd.g 𝐺 = ( Scalar ‘ 𝐿 )
asclpropd.1 ( 𝜑𝑃 = ( Base ‘ 𝐹 ) )
asclpropd.2 ( 𝜑𝑃 = ( Base ‘ 𝐺 ) )
asclpropd.3 ( ( 𝜑 ∧ ( 𝑥𝑃𝑦𝑊 ) ) → ( 𝑥 ( ·𝑠𝐾 ) 𝑦 ) = ( 𝑥 ( ·𝑠𝐿 ) 𝑦 ) )
asclpropd.4 ( 𝜑 → ( 1r𝐾 ) = ( 1r𝐿 ) )
asclpropd.5 ( 𝜑 → ( 1r𝐾 ) ∈ 𝑊 )
Assertion asclpropd ( 𝜑 → ( algSc ‘ 𝐾 ) = ( algSc ‘ 𝐿 ) )

Proof

Step Hyp Ref Expression
1 asclpropd.f 𝐹 = ( Scalar ‘ 𝐾 )
2 asclpropd.g 𝐺 = ( Scalar ‘ 𝐿 )
3 asclpropd.1 ( 𝜑𝑃 = ( Base ‘ 𝐹 ) )
4 asclpropd.2 ( 𝜑𝑃 = ( Base ‘ 𝐺 ) )
5 asclpropd.3 ( ( 𝜑 ∧ ( 𝑥𝑃𝑦𝑊 ) ) → ( 𝑥 ( ·𝑠𝐾 ) 𝑦 ) = ( 𝑥 ( ·𝑠𝐿 ) 𝑦 ) )
6 asclpropd.4 ( 𝜑 → ( 1r𝐾 ) = ( 1r𝐿 ) )
7 asclpropd.5 ( 𝜑 → ( 1r𝐾 ) ∈ 𝑊 )
8 5 oveqrspc2v ( ( 𝜑 ∧ ( 𝑧𝑃 ∧ ( 1r𝐾 ) ∈ 𝑊 ) ) → ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐾 ) ) )
9 8 anassrs ( ( ( 𝜑𝑧𝑃 ) ∧ ( 1r𝐾 ) ∈ 𝑊 ) → ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐾 ) ) )
10 7 9 mpidan ( ( 𝜑𝑧𝑃 ) → ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐾 ) ) )
11 6 oveq2d ( 𝜑 → ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) )
12 11 adantr ( ( 𝜑𝑧𝑃 ) → ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) )
13 10 12 eqtrd ( ( 𝜑𝑧𝑃 ) → ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) = ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) )
14 13 mpteq2dva ( 𝜑 → ( 𝑧𝑃 ↦ ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) ) = ( 𝑧𝑃 ↦ ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) ) )
15 3 mpteq1d ( 𝜑 → ( 𝑧𝑃 ↦ ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) ) = ( 𝑧 ∈ ( Base ‘ 𝐹 ) ↦ ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) ) )
16 4 mpteq1d ( 𝜑 → ( 𝑧𝑃 ↦ ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) ) = ( 𝑧 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) ) )
17 14 15 16 3eqtr3d ( 𝜑 → ( 𝑧 ∈ ( Base ‘ 𝐹 ) ↦ ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) ) = ( 𝑧 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) ) )
18 eqid ( algSc ‘ 𝐾 ) = ( algSc ‘ 𝐾 )
19 eqid ( Base ‘ 𝐹 ) = ( Base ‘ 𝐹 )
20 eqid ( ·𝑠𝐾 ) = ( ·𝑠𝐾 )
21 eqid ( 1r𝐾 ) = ( 1r𝐾 )
22 18 1 19 20 21 asclfval ( algSc ‘ 𝐾 ) = ( 𝑧 ∈ ( Base ‘ 𝐹 ) ↦ ( 𝑧 ( ·𝑠𝐾 ) ( 1r𝐾 ) ) )
23 eqid ( algSc ‘ 𝐿 ) = ( algSc ‘ 𝐿 )
24 eqid ( Base ‘ 𝐺 ) = ( Base ‘ 𝐺 )
25 eqid ( ·𝑠𝐿 ) = ( ·𝑠𝐿 )
26 eqid ( 1r𝐿 ) = ( 1r𝐿 )
27 23 2 24 25 26 asclfval ( algSc ‘ 𝐿 ) = ( 𝑧 ∈ ( Base ‘ 𝐺 ) ↦ ( 𝑧 ( ·𝑠𝐿 ) ( 1r𝐿 ) ) )
28 17 22 27 3eqtr4g ( 𝜑 → ( algSc ‘ 𝐾 ) = ( algSc ‘ 𝐿 ) )