MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f18(a, b, c, d, e, f, g, h, i, j) -> f24(a, b, c, d, e, f, g, h, i, j) [ 0 >= a ]
		(?, 1)    f31(a, b, c, d, e, f, g, h, i, j) -> f31(a, b, c, d, e, f, g, h, i, j)
		(?, 1)    f33(a, b, c, d, e, f, g, h, i, j) -> f36(a, b, c, d, e, f, g, h, i, j)
		(?, 1)    f24(a, b, c, d, e, f, g, h, i, j) -> f31(a, k, c, d, e, f, g, h, i, j) [ a >= 1 ]
		(?, 1)    f24(a, b, c, d, e, f, g, h, i, j) -> f31(a, l, c, k, e, f, g, h, i, j) [ 0 >= a /\ c + 999 >= k ]
		(?, 1)    f24(a, b, c, d, e, f, g, h, i, j) -> f31(1, l, c, k, e, f, g, h, i, j) [ 0 >= a /\ k >= c + 1000 ]
		(?, 1)    f18(a, b, c, d, e, f, g, h, i, j) -> f24(0, b, k, d, k, f, g, h, i, j) [ a >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j) -> f18(1, b, c, d, e, k, k, h, i, j) [ 0 >= k ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j) -> f18(1, b, c, d, e, k, 0, 1, l, l) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j) -> f31(1, b, c, d, e, k, 0, 1, l, l) [ k >= 1 /\ 0 >= l ]
	start location:	f0
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [a, c].
We thus obtain the following problem:
2:	T:
		(1, 1)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(?, 1)    f18(a, c) -> f24(0, k) [ a >= 1 ]
		(?, 1)    f24(a, c) -> f31(1, c) [ 0 >= a /\ k >= c + 1000 ]
		(?, 1)    f24(a, c) -> f31(a, c) [ 0 >= a /\ c + 999 >= k ]
		(?, 1)    f24(a, c) -> f31(a, c) [ a >= 1 ]
		(?, 1)    f33(a, c) -> f36(a, c)
		(?, 1)    f31(a, c) -> f31(a, c)
		(?, 1)    f18(a, c) -> f24(a, c) [ 0 >= a ]
	start location:	f0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 2 produces the following problem:
3:	T:
		(1, 1)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(?, 1)    f18(a, c) -> f24(0, k) [ a >= 1 ]
		(?, 1)    f24(a, c) -> f31(1, c) [ 0 >= a /\ k >= c + 1000 ]
		(?, 1)    f24(a, c) -> f31(a, c) [ 0 >= a /\ c + 999 >= k ]
		(?, 1)    f24(a, c) -> f31(a, c) [ a >= 1 ]
		(?, 1)    f31(a, c) -> f31(a, c)
		(?, 1)    f18(a, c) -> f24(a, c) [ 0 >= a ]
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transitions from problem 3:
	f24(a, c) -> f31(a, c) [ a >= 1 ]
	f18(a, c) -> f24(a, c) [ 0 >= a ]
We thus obtain the following problem:
4:	T:
		(?, 1)    f24(a, c) -> f31(a, c) [ 0 >= a /\ c + 999 >= k ]
		(?, 1)    f24(a, c) -> f31(1, c) [ 0 >= a /\ k >= c + 1000 ]
		(?, 1)    f18(a, c) -> f24(0, k) [ a >= 1 ]
		(?, 1)    f31(a, c) -> f31(a, c)
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ]
	start location:	f0
	leaf cost:	1

Repeatedly propagating knowledge in problem 4 produces the following problem:
5:	T:
		(2, 1)    f24(a, c) -> f31(a, c) [ 0 >= a /\ c + 999 >= k ]
		(2, 1)    f24(a, c) -> f31(1, c) [ 0 >= a /\ k >= c + 1000 ]
		(2, 1)    f18(a, c) -> f24(0, k) [ a >= 1 ]
		(?, 1)    f31(a, c) -> f31(a, c)
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ]
	start location:	f0
	leaf cost:	1

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f18: -X_1 + 1 >= 0 /\ X_1 - 1 >= 0
  For symbol f24: -X_1 >= 0 /\ X_1 >= 0
  For symbol f31: -X_1 + 1 >= 0 /\ X_1 >= 0


This yielded the following problem:
6:	T:
		(1, 1)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(?, 1)    f31(a, c) -> f31(a, c) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f18(a, c) -> f24(0, k) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f24(a, c) -> f31(1, c) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ k >= c + 1000 ]
		(2, 1)    f24(a, c) -> f31(a, c) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ c + 999 >= k ]
	start location:	f0
	leaf cost:	1

By chaining the transition f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l ] with all transitions in problem 6, the following new transition is obtained:
	f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l /\ 0 >= 0 /\ 1 >= 0 ]
We thus obtain the following problem:
7:	T:
		(1, 2)    f0(a, c) -> f31(1, c) [ k >= 1 /\ 0 >= l /\ 0 >= 0 /\ 1 >= 0 ]
		(1, 1)    f0(a, c) -> f18(1, c) [ k >= 1 /\ l >= 1 ]
		(1, 1)    f0(a, c) -> f18(1, c) [ 0 >= k ]
		(?, 1)    f31(a, c) -> f31(a, c) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f18(a, c) -> f24(0, k) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f24(a, c) -> f31(1, c) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ k >= c + 1000 ]
		(2, 1)    f24(a, c) -> f31(a, c) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ c + 999 >= k ]
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 0.403 sec (SMT: 0.380 sec)