Metamath Proof Explorer


Theorem pws0g

Description: Zero in a structure power of a monoid. (Contributed by Mario Carneiro, 11-Jan-2015)

Ref Expression
Hypotheses pwsmnd.y 𝑌 = ( 𝑅s 𝐼 )
pws0g.z 0 = ( 0g𝑅 )
Assertion pws0g ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 𝐼 × { 0 } ) = ( 0g𝑌 ) )

Proof

Step Hyp Ref Expression
1 pwsmnd.y 𝑌 = ( 𝑅s 𝐼 )
2 pws0g.z 0 = ( 0g𝑅 )
3 eqid ( ( Scalar ‘ 𝑅 ) Xs ( 𝐼 × { 𝑅 } ) ) = ( ( Scalar ‘ 𝑅 ) Xs ( 𝐼 × { 𝑅 } ) )
4 simpr ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → 𝐼𝑉 )
5 fvexd ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( Scalar ‘ 𝑅 ) ∈ V )
6 fconst6g ( 𝑅 ∈ Mnd → ( 𝐼 × { 𝑅 } ) : 𝐼 ⟶ Mnd )
7 6 adantr ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 𝐼 × { 𝑅 } ) : 𝐼 ⟶ Mnd )
8 3 4 5 7 prds0g ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 0g ∘ ( 𝐼 × { 𝑅 } ) ) = ( 0g ‘ ( ( Scalar ‘ 𝑅 ) Xs ( 𝐼 × { 𝑅 } ) ) ) )
9 fconstmpt ( 𝐼 × { 0 } ) = ( 𝑥𝐼0 )
10 elex ( 𝑅 ∈ Mnd → 𝑅 ∈ V )
11 10 ad2antrr ( ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) ∧ 𝑥𝐼 ) → 𝑅 ∈ V )
12 fconstmpt ( 𝐼 × { 𝑅 } ) = ( 𝑥𝐼𝑅 )
13 12 a1i ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 𝐼 × { 𝑅 } ) = ( 𝑥𝐼𝑅 ) )
14 fn0g 0g Fn V
15 14 a1i ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → 0g Fn V )
16 dffn5 ( 0g Fn V ↔ 0g = ( 𝑟 ∈ V ↦ ( 0g𝑟 ) ) )
17 15 16 sylib ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → 0g = ( 𝑟 ∈ V ↦ ( 0g𝑟 ) ) )
18 fveq2 ( 𝑟 = 𝑅 → ( 0g𝑟 ) = ( 0g𝑅 ) )
19 18 2 eqtr4di ( 𝑟 = 𝑅 → ( 0g𝑟 ) = 0 )
20 11 13 17 19 fmptco ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 0g ∘ ( 𝐼 × { 𝑅 } ) ) = ( 𝑥𝐼0 ) )
21 9 20 eqtr4id ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 𝐼 × { 0 } ) = ( 0g ∘ ( 𝐼 × { 𝑅 } ) ) )
22 eqid ( Scalar ‘ 𝑅 ) = ( Scalar ‘ 𝑅 )
23 1 22 pwsval ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → 𝑌 = ( ( Scalar ‘ 𝑅 ) Xs ( 𝐼 × { 𝑅 } ) ) )
24 23 fveq2d ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 0g𝑌 ) = ( 0g ‘ ( ( Scalar ‘ 𝑅 ) Xs ( 𝐼 × { 𝑅 } ) ) ) )
25 8 21 24 3eqtr4d ( ( 𝑅 ∈ Mnd ∧ 𝐼𝑉 ) → ( 𝐼 × { 0 } ) = ( 0g𝑌 ) )