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Metamath Proof Explorer

Theorem fsumdifsnconst

Description: The sum of constant terms ( k is not free in C ) over an index set excluding a singleton. (Contributed by AV, 7-Jan-2022)

Ref Expression
Assertion fsumdifsnconst ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → Σ 𝑘 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝐶 = ( ( ( ♯ ‘ 𝐴 ) − 1 ) · 𝐶 ) )

Proof

Step Hyp Ref Expression
1 diffi ⊢ ( 𝐴 ∈ Fin → ( 𝐴 ∖ { 𝐵 } ) ∈ Fin )
2 1 anim1i ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 ∖ { 𝐵 } ) ∈ Fin ∧ 𝐶 ∈ ℂ ) )
3 2 3adant2 ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → ( ( 𝐴 ∖ { 𝐵 } ) ∈ Fin ∧ 𝐶 ∈ ℂ ) )
4 fsumconst ⊢ ( ( ( 𝐴 ∖ { 𝐵 } ) ∈ Fin ∧ 𝐶 ∈ ℂ ) → Σ 𝑘 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝐶 = ( ( ♯ ‘ ( 𝐴 ∖ { 𝐵 } ) ) · 𝐶 ) )
5 3 4 syl ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → Σ 𝑘 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝐶 = ( ( ♯ ‘ ( 𝐴 ∖ { 𝐵 } ) ) · 𝐶 ) )
6 hashdifsn ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ) → ( ♯ ‘ ( 𝐴 ∖ { 𝐵 } ) ) = ( ( ♯ ‘ 𝐴 ) − 1 ) )
7 6 3adant3 ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → ( ♯ ‘ ( 𝐴 ∖ { 𝐵 } ) ) = ( ( ♯ ‘ 𝐴 ) − 1 ) )
8 7 oveq1d ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → ( ( ♯ ‘ ( 𝐴 ∖ { 𝐵 } ) ) · 𝐶 ) = ( ( ( ♯ ‘ 𝐴 ) − 1 ) · 𝐶 ) )
9 5 8 eqtrd ⊢ ( ( 𝐴 ∈ Fin ∧ 𝐵 ∈ 𝐴 ∧ 𝐶 ∈ ℂ ) → Σ 𝑘 ∈ ( 𝐴 ∖ { 𝐵 } ) 𝐶 = ( ( ( ♯ ‘ 𝐴 ) − 1 ) · 𝐶 ) )