... xn +1 .Now∅=I⊆In +1 ( 1) |I| 1 vi∈Ixi=∅=I⊆In( 1) |I| 1 vi∈Ixi+I⊆In +1 n,n +1 I( 1) |I| 1 vi∈Ixi+I⊆In +1 n∈I,n +1 I( 1) |I| 1 vi∈Ixi.22We claim thatI⊆In +1 n,n +1 I( 1) |I| 1 vi∈Ixi+I⊆In +1 n∈I,n +1 I( 1) |I| 1 vi∈Ixi=0 and ... following event structures E 1 = E 1 , ≤, #, E4= E4, ≤, # where E4is defined as follows:• E4= { a 1 ,a2,b 1 ,b2,c 1 ,c2,d 1 ,d2,e 1 ,e2};• e 1 ≤a 1 ,b 1 ,c 1 ,d 1 ,e2≤a2,b2,c2,d2;• ... thatI⊆In +1 n,n +1 I( 1) |I| 1 vi∈Ixi+I⊆In +1 n∈I,n +1 I( 1) |I| 1 vi∈Ixi=0 and this would prove our lemma. To prove the claimI⊆In +1 n,n +1 I( 1) |I| 1 vi∈Ixi=I⊆In 1 ( 1) |I| 1 vi∈Ixi∨ xn∨ xn +1 =I⊆In 1 ( 1) |I| 1 vi∈Ixi∨...