Metamath Proof Explorer
Description: The scalar product of a monoid ring. (Contributed by Rohan Ridenour, 14-May-2024) (Proof shortened by AV, 1-Nov-2024)
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Ref |
Expression |
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Hypotheses |
mnringvscad.1 |
⊢ 𝐹 = ( 𝑅 MndRing 𝑀 ) |
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mnringvscad.2 |
⊢ 𝐵 = ( Base ‘ 𝑀 ) |
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mnringvscad.3 |
⊢ 𝑉 = ( 𝑅 freeLMod 𝐵 ) |
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mnringvscad.4 |
⊢ ( 𝜑 → 𝑅 ∈ 𝑈 ) |
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mnringvscad.5 |
⊢ ( 𝜑 → 𝑀 ∈ 𝑊 ) |
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Assertion |
mnringvscad |
⊢ ( 𝜑 → ( ·𝑠 ‘ 𝑉 ) = ( ·𝑠 ‘ 𝐹 ) ) |
Proof
Step |
Hyp |
Ref |
Expression |
1 |
|
mnringvscad.1 |
⊢ 𝐹 = ( 𝑅 MndRing 𝑀 ) |
2 |
|
mnringvscad.2 |
⊢ 𝐵 = ( Base ‘ 𝑀 ) |
3 |
|
mnringvscad.3 |
⊢ 𝑉 = ( 𝑅 freeLMod 𝐵 ) |
4 |
|
mnringvscad.4 |
⊢ ( 𝜑 → 𝑅 ∈ 𝑈 ) |
5 |
|
mnringvscad.5 |
⊢ ( 𝜑 → 𝑀 ∈ 𝑊 ) |
6 |
|
vscaid |
⊢ ·𝑠 = Slot ( ·𝑠 ‘ ndx ) |
7 |
|
vscandxnmulrndx |
⊢ ( ·𝑠 ‘ ndx ) ≠ ( .r ‘ ndx ) |
8 |
1 6 7 2 3 4 5
|
mnringnmulrd |
⊢ ( 𝜑 → ( ·𝑠 ‘ 𝑉 ) = ( ·𝑠 ‘ 𝐹 ) ) |