Metamath Proof Explorer


Theorem prodtp

Description: A product over a triple is the product of the elements. (Contributed by Thierry Arnoux, 1-Jan-2022)

Ref Expression
Hypotheses prodpr.1 ( 𝑘 = 𝐴𝐷 = 𝐸 )
prodpr.2 ( 𝑘 = 𝐵𝐷 = 𝐹 )
prodpr.a ( 𝜑𝐴𝑉 )
prodpr.b ( 𝜑𝐵𝑊 )
prodpr.e ( 𝜑𝐸 ∈ ℂ )
prodpr.f ( 𝜑𝐹 ∈ ℂ )
prodpr.3 ( 𝜑𝐴𝐵 )
prodtp.1 ( 𝑘 = 𝐶𝐷 = 𝐺 )
prodtp.c ( 𝜑𝐶𝑋 )
prodtp.g ( 𝜑𝐺 ∈ ℂ )
prodtp.2 ( 𝜑𝐴𝐶 )
prodtp.3 ( 𝜑𝐵𝐶 )
Assertion prodtp ( 𝜑 → ∏ 𝑘 ∈ { 𝐴 , 𝐵 , 𝐶 } 𝐷 = ( ( 𝐸 · 𝐹 ) · 𝐺 ) )

Proof

Step Hyp Ref Expression
1 prodpr.1 ( 𝑘 = 𝐴𝐷 = 𝐸 )
2 prodpr.2 ( 𝑘 = 𝐵𝐷 = 𝐹 )
3 prodpr.a ( 𝜑𝐴𝑉 )
4 prodpr.b ( 𝜑𝐵𝑊 )
5 prodpr.e ( 𝜑𝐸 ∈ ℂ )
6 prodpr.f ( 𝜑𝐹 ∈ ℂ )
7 prodpr.3 ( 𝜑𝐴𝐵 )
8 prodtp.1 ( 𝑘 = 𝐶𝐷 = 𝐺 )
9 prodtp.c ( 𝜑𝐶𝑋 )
10 prodtp.g ( 𝜑𝐺 ∈ ℂ )
11 prodtp.2 ( 𝜑𝐴𝐶 )
12 prodtp.3 ( 𝜑𝐵𝐶 )
13 disjprsn ( ( 𝐴𝐶𝐵𝐶 ) → ( { 𝐴 , 𝐵 } ∩ { 𝐶 } ) = ∅ )
14 11 12 13 syl2anc ( 𝜑 → ( { 𝐴 , 𝐵 } ∩ { 𝐶 } ) = ∅ )
15 df-tp { 𝐴 , 𝐵 , 𝐶 } = ( { 𝐴 , 𝐵 } ∪ { 𝐶 } )
16 15 a1i ( 𝜑 → { 𝐴 , 𝐵 , 𝐶 } = ( { 𝐴 , 𝐵 } ∪ { 𝐶 } ) )
17 tpfi { 𝐴 , 𝐵 , 𝐶 } ∈ Fin
18 17 a1i ( 𝜑 → { 𝐴 , 𝐵 , 𝐶 } ∈ Fin )
19 vex 𝑘 ∈ V
20 19 eltp ( 𝑘 ∈ { 𝐴 , 𝐵 , 𝐶 } ↔ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) )
21 1 adantl ( ( 𝜑𝑘 = 𝐴 ) → 𝐷 = 𝐸 )
22 5 adantr ( ( 𝜑𝑘 = 𝐴 ) → 𝐸 ∈ ℂ )
23 21 22 eqeltrd ( ( 𝜑𝑘 = 𝐴 ) → 𝐷 ∈ ℂ )
24 23 adantlr ( ( ( 𝜑 ∧ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) ) ∧ 𝑘 = 𝐴 ) → 𝐷 ∈ ℂ )
25 2 adantl ( ( 𝜑𝑘 = 𝐵 ) → 𝐷 = 𝐹 )
26 6 adantr ( ( 𝜑𝑘 = 𝐵 ) → 𝐹 ∈ ℂ )
27 25 26 eqeltrd ( ( 𝜑𝑘 = 𝐵 ) → 𝐷 ∈ ℂ )
28 27 adantlr ( ( ( 𝜑 ∧ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) ) ∧ 𝑘 = 𝐵 ) → 𝐷 ∈ ℂ )
29 8 adantl ( ( 𝜑𝑘 = 𝐶 ) → 𝐷 = 𝐺 )
30 10 adantr ( ( 𝜑𝑘 = 𝐶 ) → 𝐺 ∈ ℂ )
31 29 30 eqeltrd ( ( 𝜑𝑘 = 𝐶 ) → 𝐷 ∈ ℂ )
32 31 adantlr ( ( ( 𝜑 ∧ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) ) ∧ 𝑘 = 𝐶 ) → 𝐷 ∈ ℂ )
33 simpr ( ( 𝜑 ∧ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) ) → ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) )
34 24 28 32 33 mpjao3dan ( ( 𝜑 ∧ ( 𝑘 = 𝐴𝑘 = 𝐵𝑘 = 𝐶 ) ) → 𝐷 ∈ ℂ )
35 20 34 sylan2b ( ( 𝜑𝑘 ∈ { 𝐴 , 𝐵 , 𝐶 } ) → 𝐷 ∈ ℂ )
36 14 16 18 35 fprodsplit ( 𝜑 → ∏ 𝑘 ∈ { 𝐴 , 𝐵 , 𝐶 } 𝐷 = ( ∏ 𝑘 ∈ { 𝐴 , 𝐵 } 𝐷 · ∏ 𝑘 ∈ { 𝐶 } 𝐷 ) )
37 1 2 3 4 5 6 7 prodpr ( 𝜑 → ∏ 𝑘 ∈ { 𝐴 , 𝐵 } 𝐷 = ( 𝐸 · 𝐹 ) )
38 8 prodsn ( ( 𝐶𝑋𝐺 ∈ ℂ ) → ∏ 𝑘 ∈ { 𝐶 } 𝐷 = 𝐺 )
39 9 10 38 syl2anc ( 𝜑 → ∏ 𝑘 ∈ { 𝐶 } 𝐷 = 𝐺 )
40 37 39 oveq12d ( 𝜑 → ( ∏ 𝑘 ∈ { 𝐴 , 𝐵 } 𝐷 · ∏ 𝑘 ∈ { 𝐶 } 𝐷 ) = ( ( 𝐸 · 𝐹 ) · 𝐺 ) )
41 36 40 eqtrd ( 𝜑 → ∏ 𝑘 ∈ { 𝐴 , 𝐵 , 𝐶 } 𝐷 = ( ( 𝐸 · 𝐹 ) · 𝐺 ) )