Metamath Proof Explorer


Theorem psrlinv

Description: The negative function in the ring of power series. (Contributed by Mario Carneiro, 29-Dec-2014)

Ref Expression
Hypotheses psrgrp.s 𝑆 = ( 𝐼 mPwSer 𝑅 )
psrgrp.i ( 𝜑𝐼𝑉 )
psrgrp.r ( 𝜑𝑅 ∈ Grp )
psrnegcl.d 𝐷 = { 𝑓 ∈ ( ℕ0m 𝐼 ) ∣ ( 𝑓 “ ℕ ) ∈ Fin }
psrnegcl.i 𝑁 = ( invg𝑅 )
psrnegcl.b 𝐵 = ( Base ‘ 𝑆 )
psrnegcl.z ( 𝜑𝑋𝐵 )
psrlinv.o 0 = ( 0g𝑅 )
psrlinv.p + = ( +g𝑆 )
Assertion psrlinv ( 𝜑 → ( ( 𝑁𝑋 ) + 𝑋 ) = ( 𝐷 × { 0 } ) )

Proof

Step Hyp Ref Expression
1 psrgrp.s 𝑆 = ( 𝐼 mPwSer 𝑅 )
2 psrgrp.i ( 𝜑𝐼𝑉 )
3 psrgrp.r ( 𝜑𝑅 ∈ Grp )
4 psrnegcl.d 𝐷 = { 𝑓 ∈ ( ℕ0m 𝐼 ) ∣ ( 𝑓 “ ℕ ) ∈ Fin }
5 psrnegcl.i 𝑁 = ( invg𝑅 )
6 psrnegcl.b 𝐵 = ( Base ‘ 𝑆 )
7 psrnegcl.z ( 𝜑𝑋𝐵 )
8 psrlinv.o 0 = ( 0g𝑅 )
9 psrlinv.p + = ( +g𝑆 )
10 ovex ( ℕ0m 𝐼 ) ∈ V
11 4 10 rabex2 𝐷 ∈ V
12 11 a1i ( 𝜑𝐷 ∈ V )
13 fvexd ( ( 𝜑𝑥𝐷 ) → ( 𝑁 ‘ ( 𝑋𝑥 ) ) ∈ V )
14 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
15 1 14 4 6 7 psrelbas ( 𝜑𝑋 : 𝐷 ⟶ ( Base ‘ 𝑅 ) )
16 15 ffvelrnda ( ( 𝜑𝑥𝐷 ) → ( 𝑋𝑥 ) ∈ ( Base ‘ 𝑅 ) )
17 15 feqmptd ( 𝜑𝑋 = ( 𝑥𝐷 ↦ ( 𝑋𝑥 ) ) )
18 14 5 3 grpinvf1o ( 𝜑𝑁 : ( Base ‘ 𝑅 ) –1-1-onto→ ( Base ‘ 𝑅 ) )
19 f1of ( 𝑁 : ( Base ‘ 𝑅 ) –1-1-onto→ ( Base ‘ 𝑅 ) → 𝑁 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑅 ) )
20 18 19 syl ( 𝜑𝑁 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑅 ) )
21 20 feqmptd ( 𝜑𝑁 = ( 𝑦 ∈ ( Base ‘ 𝑅 ) ↦ ( 𝑁𝑦 ) ) )
22 fveq2 ( 𝑦 = ( 𝑋𝑥 ) → ( 𝑁𝑦 ) = ( 𝑁 ‘ ( 𝑋𝑥 ) ) )
23 16 17 21 22 fmptco ( 𝜑 → ( 𝑁𝑋 ) = ( 𝑥𝐷 ↦ ( 𝑁 ‘ ( 𝑋𝑥 ) ) ) )
24 12 13 16 23 17 offval2 ( 𝜑 → ( ( 𝑁𝑋 ) ∘f ( +g𝑅 ) 𝑋 ) = ( 𝑥𝐷 ↦ ( ( 𝑁 ‘ ( 𝑋𝑥 ) ) ( +g𝑅 ) ( 𝑋𝑥 ) ) ) )
25 eqid ( +g𝑅 ) = ( +g𝑅 )
26 1 2 3 4 5 6 7 psrnegcl ( 𝜑 → ( 𝑁𝑋 ) ∈ 𝐵 )
27 1 6 25 9 26 7 psradd ( 𝜑 → ( ( 𝑁𝑋 ) + 𝑋 ) = ( ( 𝑁𝑋 ) ∘f ( +g𝑅 ) 𝑋 ) )
28 fconstmpt ( 𝐷 × { 0 } ) = ( 𝑥𝐷0 )
29 14 25 8 5 grplinv ( ( 𝑅 ∈ Grp ∧ ( 𝑋𝑥 ) ∈ ( Base ‘ 𝑅 ) ) → ( ( 𝑁 ‘ ( 𝑋𝑥 ) ) ( +g𝑅 ) ( 𝑋𝑥 ) ) = 0 )
30 3 16 29 syl2an2r ( ( 𝜑𝑥𝐷 ) → ( ( 𝑁 ‘ ( 𝑋𝑥 ) ) ( +g𝑅 ) ( 𝑋𝑥 ) ) = 0 )
31 30 mpteq2dva ( 𝜑 → ( 𝑥𝐷 ↦ ( ( 𝑁 ‘ ( 𝑋𝑥 ) ) ( +g𝑅 ) ( 𝑋𝑥 ) ) ) = ( 𝑥𝐷0 ) )
32 28 31 eqtr4id ( 𝜑 → ( 𝐷 × { 0 } ) = ( 𝑥𝐷 ↦ ( ( 𝑁 ‘ ( 𝑋𝑥 ) ) ( +g𝑅 ) ( 𝑋𝑥 ) ) ) )
33 24 27 32 3eqtr4d ( 𝜑 → ( ( 𝑁𝑋 ) + 𝑋 ) = ( 𝐷 × { 0 } ) )