Metamath Proof Explorer


Theorem cndprobin

Description: An identity linking conditional probability and intersection. (Contributed by Thierry Arnoux, 13-Dec-2016) (Revised by Thierry Arnoux, 21-Jan-2017)

Ref Expression
Assertion cndprobin ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( ( ( cprob ‘ 𝑃 ) ‘ ⟨ 𝐴 , 𝐵 ⟩ ) · ( 𝑃 ‘ 𝐵 ) ) = ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) )

Proof

Step Hyp Ref Expression
1 cndprobval ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( ( cprob ‘ 𝑃 ) ‘ ⟨ 𝐴 , 𝐵 ⟩ ) = ( ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) / ( 𝑃 ‘ 𝐵 ) ) )
2 1 oveq1d ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( ( ( cprob ‘ 𝑃 ) ‘ ⟨ 𝐴 , 𝐵 ⟩ ) · ( 𝑃 ‘ 𝐵 ) ) = ( ( ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) / ( 𝑃 ‘ 𝐵 ) ) · ( 𝑃 ‘ 𝐵 ) ) )
3 2 adantr ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( ( ( cprob ‘ 𝑃 ) ‘ ⟨ 𝐴 , 𝐵 ⟩ ) · ( 𝑃 ‘ 𝐵 ) ) = ( ( ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) / ( 𝑃 ‘ 𝐵 ) ) · ( 𝑃 ‘ 𝐵 ) ) )
4 unitsscn ⊢ ( 0 [,] 1 ) ⊆ ℂ
5 simp1 ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → 𝑃 ∈ Prob )
6 domprobsiga ⊢ ( 𝑃 ∈ Prob → dom 𝑃 ∈ ∪ ran sigAlgebra )
7 inelsiga ⊢ ( ( dom 𝑃 ∈ ∪ ran sigAlgebra ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝐴 ∩ 𝐵 ) ∈ dom 𝑃 )
8 6 7 syl3an1 ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝐴 ∩ 𝐵 ) ∈ dom 𝑃 )
9 prob01 ⊢ ( ( 𝑃 ∈ Prob ∧ ( 𝐴 ∩ 𝐵 ) ∈ dom 𝑃 ) → ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) ∈ ( 0 [,] 1 ) )
10 5 8 9 syl2anc ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) ∈ ( 0 [,] 1 ) )
11 4 10 sselid ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) ∈ ℂ )
12 11 adantr ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) ∈ ℂ )
13 prob01 ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝑃 ‘ 𝐵 ) ∈ ( 0 [,] 1 ) )
14 13 3adant2 ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝑃 ‘ 𝐵 ) ∈ ( 0 [,] 1 ) )
15 4 14 sselid ⊢ ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) → ( 𝑃 ‘ 𝐵 ) ∈ ℂ )
16 15 adantr ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( 𝑃 ‘ 𝐵 ) ∈ ℂ )
17 simpr ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( 𝑃 ‘ 𝐵 ) ≠ 0 )
18 12 16 17 divcan1d ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( ( ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) / ( 𝑃 ‘ 𝐵 ) ) · ( 𝑃 ‘ 𝐵 ) ) = ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) )
19 3 18 eqtrd ⊢ ( ( ( 𝑃 ∈ Prob ∧ 𝐴 ∈ dom 𝑃 ∧ 𝐵 ∈ dom 𝑃 ) ∧ ( 𝑃 ‘ 𝐵 ) ≠ 0 ) → ( ( ( cprob ‘ 𝑃 ) ‘ ⟨ 𝐴 , 𝐵 ⟩ ) · ( 𝑃 ‘ 𝐵 ) ) = ( 𝑃 ‘ ( 𝐴 ∩ 𝐵 ) ) )