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Metamath Proof Explorer

Theorem shftval2

Description: Value of a sequence shifted by A - B . (Contributed by NM, 20-Jul-2005) (Revised by Mario Carneiro, 5-Nov-2013)

Ref Expression
Hypothesis shftfval.1 ⊢ 𝐹 ∈ V
Assertion shftval2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐹 shift ( 𝐴 − 𝐵 ) ) ‘ ( 𝐴 + 𝐶 ) ) = ( 𝐹 ‘ ( 𝐵 + 𝐶 ) ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 ⊢ 𝐹 ∈ V
2 subcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
3 2 3adant3 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 − 𝐵 ) ∈ ℂ )
4 addcl ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐴 + 𝐶 ) ∈ ℂ )
5 1 shftval ⊢ ( ( ( 𝐴 − 𝐵 ) ∈ ℂ ∧ ( 𝐴 + 𝐶 ) ∈ ℂ ) → ( ( 𝐹 shift ( 𝐴 − 𝐵 ) ) ‘ ( 𝐴 + 𝐶 ) ) = ( 𝐹 ‘ ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) ) )
6 3 4 5 3imp3i2an ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐹 shift ( 𝐴 − 𝐵 ) ) ‘ ( 𝐴 + 𝐶 ) ) = ( 𝐹 ‘ ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) ) )
7 pnncan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐶 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) = ( 𝐶 + 𝐵 ) )
8 7 3com23 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) = ( 𝐶 + 𝐵 ) )
9 addcom ⊢ ( ( 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
10 9 3adant1 ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐵 + 𝐶 ) = ( 𝐶 + 𝐵 ) )
11 8 10 eqtr4d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) = ( 𝐵 + 𝐶 ) )
12 11 fveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( 𝐹 ‘ ( ( 𝐴 + 𝐶 ) − ( 𝐴 − 𝐵 ) ) ) = ( 𝐹 ‘ ( 𝐵 + 𝐶 ) ) )
13 6 12 eqtrd ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ∧ 𝐶 ∈ ℂ ) → ( ( 𝐹 shift ( 𝐴 − 𝐵 ) ) ‘ ( 𝐴 + 𝐶 ) ) = ( 𝐹 ‘ ( 𝐵 + 𝐶 ) ) )