1# Check that isl recognizes that the inverse schedule is single-valued 2# and does not end up in an infinite recursion. 3[t1] -> {S[c2] -> [c2]: t1 <= c2 <= 134 and (c2+t1) % 128 = 0 and c2 > 0} 4[t1] -> {: t1 > 0} 5[t1] -> {} 6