Metamath Proof Explorer


Theorem psdval

Description: Evaluate the partial derivative of a power series. (Contributed by SN, 11-Apr-2025)

Ref Expression
Hypotheses psdffval.s 𝑆 = ( 𝐼 mPwSer 𝑅 )
psdffval.b 𝐵 = ( Base ‘ 𝑆 )
psdffval.d 𝐷 = { ∈ ( ℕ0m 𝐼 ) ∣ ( “ ℕ ) ∈ Fin }
psdffval.i ( 𝜑𝐼𝑉 )
psdffval.r ( 𝜑𝑅𝑊 )
psdfval.x ( 𝜑𝑋𝐼 )
psdval.f ( 𝜑𝐹𝐵 )
Assertion psdval ( 𝜑 → ( ( ( 𝐼 mPSDer 𝑅 ) ‘ 𝑋 ) ‘ 𝐹 ) = ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) )

Proof

Step Hyp Ref Expression
1 psdffval.s 𝑆 = ( 𝐼 mPwSer 𝑅 )
2 psdffval.b 𝐵 = ( Base ‘ 𝑆 )
3 psdffval.d 𝐷 = { ∈ ( ℕ0m 𝐼 ) ∣ ( “ ℕ ) ∈ Fin }
4 psdffval.i ( 𝜑𝐼𝑉 )
5 psdffval.r ( 𝜑𝑅𝑊 )
6 psdfval.x ( 𝜑𝑋𝐼 )
7 psdval.f ( 𝜑𝐹𝐵 )
8 1 2 3 4 5 6 psdfval ( 𝜑 → ( ( 𝐼 mPSDer 𝑅 ) ‘ 𝑋 ) = ( 𝑓𝐵 ↦ ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝑓 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) ) )
9 fveq1 ( 𝑓 = 𝐹 → ( 𝑓 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) = ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) )
10 9 oveq2d ( 𝑓 = 𝐹 → ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝑓 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) = ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) )
11 10 mpteq2dv ( 𝑓 = 𝐹 → ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝑓 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) = ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) )
12 11 adantl ( ( 𝜑𝑓 = 𝐹 ) → ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝑓 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) = ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) )
13 ovex ( ℕ0m 𝐼 ) ∈ V
14 3 13 rabex2 𝐷 ∈ V
15 14 a1i ( 𝜑𝐷 ∈ V )
16 15 mptexd ( 𝜑 → ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) ∈ V )
17 8 12 7 16 fvmptd ( 𝜑 → ( ( ( 𝐼 mPSDer 𝑅 ) ‘ 𝑋 ) ‘ 𝐹 ) = ( 𝑘𝐷 ↦ ( ( ( 𝑘𝑋 ) + 1 ) ( .g𝑅 ) ( 𝐹 ‘ ( 𝑘f + ( 𝑦𝐼 ↦ if ( 𝑦 = 𝑋 , 1 , 0 ) ) ) ) ) ) )