Metamath Proof Explorer


Theorem fprodeq02

Description: If one of the factors is zero the product is zero. (Contributed by Thierry Arnoux, 11-Dec-2021)

Ref Expression
Hypotheses fprodeq02.1 ( 𝑘 = 𝐾𝐵 = 𝐶 )
fprodeq02.a ( 𝜑𝐴 ∈ Fin )
fprodeq02.b ( ( 𝜑𝑘𝐴 ) → 𝐵 ∈ ℂ )
fprodeq02.k ( 𝜑𝐾𝐴 )
fprodeq02.c ( 𝜑𝐶 = 0 )
Assertion fprodeq02 ( 𝜑 → ∏ 𝑘𝐴 𝐵 = 0 )

Proof

Step Hyp Ref Expression
1 fprodeq02.1 ( 𝑘 = 𝐾𝐵 = 𝐶 )
2 fprodeq02.a ( 𝜑𝐴 ∈ Fin )
3 fprodeq02.b ( ( 𝜑𝑘𝐴 ) → 𝐵 ∈ ℂ )
4 fprodeq02.k ( 𝜑𝐾𝐴 )
5 fprodeq02.c ( 𝜑𝐶 = 0 )
6 disjdif ( { 𝐾 } ∩ ( 𝐴 ∖ { 𝐾 } ) ) = ∅
7 6 a1i ( 𝜑 → ( { 𝐾 } ∩ ( 𝐴 ∖ { 𝐾 } ) ) = ∅ )
8 4 snssd ( 𝜑 → { 𝐾 } ⊆ 𝐴 )
9 undif ( { 𝐾 } ⊆ 𝐴 ↔ ( { 𝐾 } ∪ ( 𝐴 ∖ { 𝐾 } ) ) = 𝐴 )
10 8 9 sylib ( 𝜑 → ( { 𝐾 } ∪ ( 𝐴 ∖ { 𝐾 } ) ) = 𝐴 )
11 10 eqcomd ( 𝜑𝐴 = ( { 𝐾 } ∪ ( 𝐴 ∖ { 𝐾 } ) ) )
12 7 11 2 3 fprodsplit ( 𝜑 → ∏ 𝑘𝐴 𝐵 = ( ∏ 𝑘 ∈ { 𝐾 } 𝐵 · ∏ 𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) 𝐵 ) )
13 0cnd ( 𝜑 → 0 ∈ ℂ )
14 5 13 eqeltrd ( 𝜑𝐶 ∈ ℂ )
15 1 prodsn ( ( 𝐾𝐴𝐶 ∈ ℂ ) → ∏ 𝑘 ∈ { 𝐾 } 𝐵 = 𝐶 )
16 4 14 15 syl2anc ( 𝜑 → ∏ 𝑘 ∈ { 𝐾 } 𝐵 = 𝐶 )
17 16 5 eqtrd ( 𝜑 → ∏ 𝑘 ∈ { 𝐾 } 𝐵 = 0 )
18 17 oveq1d ( 𝜑 → ( ∏ 𝑘 ∈ { 𝐾 } 𝐵 · ∏ 𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) 𝐵 ) = ( 0 · ∏ 𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) 𝐵 ) )
19 diffi ( 𝐴 ∈ Fin → ( 𝐴 ∖ { 𝐾 } ) ∈ Fin )
20 2 19 syl ( 𝜑 → ( 𝐴 ∖ { 𝐾 } ) ∈ Fin )
21 difssd ( 𝜑 → ( 𝐴 ∖ { 𝐾 } ) ⊆ 𝐴 )
22 21 sselda ( ( 𝜑𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) ) → 𝑘𝐴 )
23 22 3 syldan ( ( 𝜑𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) ) → 𝐵 ∈ ℂ )
24 20 23 fprodcl ( 𝜑 → ∏ 𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) 𝐵 ∈ ℂ )
25 24 mul02d ( 𝜑 → ( 0 · ∏ 𝑘 ∈ ( 𝐴 ∖ { 𝐾 } ) 𝐵 ) = 0 )
26 12 18 25 3eqtrd ( 𝜑 → ∏ 𝑘𝐴 𝐵 = 0 )