#    do i = 1, np {
#	do j = 1, i {
#	    ij = ia(i) + j
#	    a = 1. / (i + j)
#	    do k = 1, i {
#		maxl = k
#		if (k == i) then
#		    maxl = j
#		endif
#		do l = 1, maxl {
#		    kl = ia(k) + l
#		    b = 1. / (k + l)
#		    val = a + b
#		    if (i == j) then
#			val = val * .5
#		    endif
#		    if (k == l) then
#			val = val * .5
#		    endif
#		    x(ij,kl) = val
#		    x(kl,ij) = val
#		}
#	    }
#	}
#    }
#! 
#
# 
# As far as I remember it was the dependence test between X(IJ,KL) and X(IJ,KL).
# 
# Pips discovered the following precondition:
# 

{[d1,d2,d3,d4] : exists ( NP,
			I,J,K,L,IJ,KJ,MAXL,
			I',J',K',L',IJ',KJ',MAXL' :
			I' = I+d1 &&
			J' = J+d2 &&
			K' = K+d3 &&
			L' = L+d4 
	&& 1<=L &&  1<=J &&  MAXL<=K &&  I<=NP &&  NP<=40 &&  10<=NP 
	&& 10+8K+MAXL<=NP+8I+J &&  38K+MAXL<=38I+J 
	&&  J+K<=I+MAXL &&  L<=MAXL
	&& 1 <= J,K <= I <= NP && 1 <=L <= MAXL

	&& 1<=L' &&  1<=J' &&  MAXL'<=K' &&  I'<=NP &&  NP<=40 &&  10<=NP 
	&& 10+8K'+MAXL'<=NP+8I'+J' &&  38K'+MAXL'<=38I'+J' 
	&&  J'+K'<=I'+MAXL' &&  L'<=MAXL'
	&& 1 <= J',K' <= I' <= NP && 1 <=L' <= MAXL'
	)};