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Metamath Proof Explorer

Theorem shftcan2

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 shftcan2 ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift - 𝐴 ) shift 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )

Proof

Step Hyp Ref Expression
1 shftfval.1 ⊢ 𝐹 ∈ V
2 negneg ⊢ ( 𝐴 ∈ ℂ → - - 𝐴 = 𝐴 )
3 2 adantr ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → - - 𝐴 = 𝐴 )
4 3 oveq2d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( 𝐹 shift - 𝐴 ) shift - - 𝐴 ) = ( ( 𝐹 shift - 𝐴 ) shift 𝐴 ) )
5 4 fveq1d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift - 𝐴 ) shift - - 𝐴 ) ‘ 𝐵 ) = ( ( ( 𝐹 shift - 𝐴 ) shift 𝐴 ) ‘ 𝐵 ) )
6 negcl ⊢ ( 𝐴 ∈ ℂ → - 𝐴 ∈ ℂ )
7 1 shftcan1 ⊢ ( ( - 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift - 𝐴 ) shift - - 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )
8 6 7 sylan ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift - 𝐴 ) shift - - 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )
9 5 8 eqtr3d ⊢ ( ( 𝐴 ∈ ℂ ∧ 𝐵 ∈ ℂ ) → ( ( ( 𝐹 shift - 𝐴 ) shift 𝐴 ) ‘ 𝐵 ) = ( 𝐹 ‘ 𝐵 ) )