By: |
|
1 |
p:(k( List)). sum(||p(j)|| | j < k) = 0 & (j:k, x,y:||p(j)||. x<y (p(j))[x]>(p(j))[y]) & (j:k, x:||p(j)||. (p(j))[x]<0 & c((p(j))[x]) = j) | 2 steps |
2 |
2. 0<n 3. k:, c:((n-1)k). 3. p:(k( List)). 3. sum(||p(j)|| | j < k) = n-1 3. & (j:k, x,y:||p(j)||. x<y (p(j))[x]>(p(j))[y]) 3. & (j:k, x:||p(j)||. (p(j))[x]<n-1 & c((p(j))[x]) = j) k:, c:(nk). p:(k( List)). sum(||p(j)|| | j < k) = n & (j:k, x,y:||p(j)||. x<y (p(j))[x]>(p(j))[y]) & (j:k, x:||p(j)||. (p(j))[x]<n & c((p(j))[x]) = j) | 23 steps |
About: