This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

Theorem shftcan1

Description: Cancellation law for the shift operation. (Contributed by NM, 4-Aug-2005) (Revised by Mario Carneiro, 5-Nov-2013)

Ref Expression
Hypothesis shftfval.1 ⊢ 𝐹 ∈ V
Assertion shftcan1 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 ⊢ 𝐹 ∈ V
2 negcl ⊢ ( 𝐴 ∈ ℂ → - 𝐴 ∈ ℂ )
3 1 2shfti ⊢ ( ( 𝐴 ∈ ℂ ∧ - 𝐴 ∈ ℂ ) → ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) = ( 𝐹 shift ( 𝐴 + - 𝐴 ) ) )
4 2 3 mpdan ⊢ ( 𝐴 ∈ ℂ → ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) = ( 𝐹 shift ( 𝐴 + - 𝐴 ) ) )
5 negid ⊢ ( 𝐴 ∈ ℂ → ( 𝐴 + - 𝐴 ) = 0 )
6 5 oveq2d ⊢ ( 𝐴 ∈ ℂ → ( 𝐹 shift ( 𝐴 + - 𝐴 ) ) = ( 𝐹 shift 0 ) )
7 4 6 eqtrd ⊢ ( 𝐴 ∈ ℂ → ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) = ( 𝐹 shift 0 ) )
8 7 fveq1d ⊢ ( 𝐴 ∈ ℂ → ( ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) ‘ 𝐵 ) = ( ( 𝐹 shift 0 ) ‘ 𝐵 ) )
9 1 shftidt ⊢ ( 𝐵 ∈ ℂ → ( ( 𝐹 shift 0 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )
10 8 9 sylan9eq ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift 𝐴 ) shift - 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )