Metamath Proof Explorer


Theorem padicval

Description: Value of the p-adic absolute value. (Contributed by Mario Carneiro, 8-Sep-2014)

Ref Expression
Hypothesis padicval.j ⊢ 𝐽 = ( 𝑞 ∈ ℙ ↦ ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑞 ↑ - ( 𝑞 pCnt 𝑥 ) ) ) ) )
Assertion padicval ( ( 𝑃 ∈ ℙ ∧ 𝑋 ∈ ℚ ) → ( ( 𝐽 ‘ 𝑃 ) ‘ 𝑋 ) = if ( 𝑋 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ) )

Proof

Step Hyp Ref Expression
1 padicval.j ⊢ 𝐽 = ( 𝑞 ∈ ℙ ↦ ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑞 ↑ - ( 𝑞 pCnt 𝑥 ) ) ) ) )
2 1 padicfval ⊢ ( 𝑃 ∈ ℙ → ( 𝐽 ‘ 𝑃 ) = ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) ) )
3 2 fveq1d ⊢ ( 𝑃 ∈ ℙ → ( ( 𝐽 ‘ 𝑃 ) ‘ 𝑋 ) = ( ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) ) ‘ 𝑋 ) )
4 eqeq1 ⊢ ( 𝑥 = 𝑋 → ( 𝑥 = 0 ↔ 𝑋 = 0 ) )
5 oveq2 ⊢ ( 𝑥 = 𝑋 → ( 𝑃 pCnt 𝑥 ) = ( 𝑃 pCnt 𝑋 ) )
6 5 negeqd ⊢ ( 𝑥 = 𝑋 → - ( 𝑃 pCnt 𝑥 ) = - ( 𝑃 pCnt 𝑋 ) )
7 6 oveq2d ⊢ ( 𝑥 = 𝑋 → ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) = ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) )
8 4 7 ifbieq2d ⊢ ( 𝑥 = 𝑋 → if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) = if ( 𝑋 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ) )
9 eqid ⊢ ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) ) = ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) )
10 c0ex ⊢ 0 ∈ V
11 ovex ⊢ ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ∈ V
12 10 11 ifex ⊢ if ( 𝑋 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ) ∈ V
13 8 9 12 fvmpt ⊢ ( 𝑋 ∈ ℚ → ( ( 𝑥 ∈ ℚ ↦ if ( 𝑥 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑥 ) ) ) ) ‘ 𝑋 ) = if ( 𝑋 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ) )
14 3 13 sylan9eq ⊢ ( ( 𝑃 ∈ ℙ ∧ 𝑋 ∈ ℚ ) → ( ( 𝐽 ‘ 𝑃 ) ‘ 𝑋 ) = if ( 𝑋 = 0 , 0 , ( 𝑃 ↑ - ( 𝑃 pCnt 𝑋 ) ) ) )