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Metamath Proof Explorer

Theorem shftval5

Description: Value of a shifted sequence. (Contributed by NM, 19-Aug-2005) (Revised by Mario Carneiro, 5-Nov-2013)

Ref Expression
Hypothesis shftfval.1 ⊢ 𝐹 ∈ V
Assertion shftval5 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) ‘ ( 𝐵 + 𝐴 ) ) = ( 𝐹 ‘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 ⊢ 𝐹 ∈ V
2 simpr ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → 𝐴 ∈ ℂ )
3 addcl ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐵 + 𝐴 ) ∈ ℂ )
4 1 shftval ⊢ ( ( 𝐴 ∈ ℂ ∧ ( 𝐵 + 𝐴 ) ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) ‘ ( 𝐵 + 𝐴 ) ) = ( 𝐹 ‘ ( ( 𝐵 + 𝐴 ) − 𝐴 ) ) )
5 2 3 4 syl2anc ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) ‘ ( 𝐵 + 𝐴 ) ) = ( 𝐹 ‘ ( ( 𝐵 + 𝐴 ) − 𝐴 ) ) )
6 pncan ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐵 + 𝐴 ) − 𝐴 ) = 𝐵 )
7 6 fveq2d ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( 𝐹 ‘ ( ( 𝐵 + 𝐴 ) − 𝐴 ) ) = ( 𝐹 ‘ 𝐵 ) )
8 5 7 eqtrd ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐴 ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) ‘ ( 𝐵 + 𝐴 ) ) = ( 𝐹 ‘ 𝐵 ) )
9 8 ancoms ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) ‘ ( 𝐵 + 𝐴 ) ) = ( 𝐹 ‘ 𝐵 ) )