MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f21(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f29(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) [ 0 >= a ]
		(?, 1)    f41(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f41(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q)
		(?, 1)    f43(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f46(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q)
		(?, 1)    f29(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f41(a, r, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) [ a >= 1 ]
		(?, 1)    f29(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f41(a, s, c, 0, r, r, r, h, i, j, k, l, m, n, o, p, q) [ 0 >= a /\ c + 999 >= r ]
		(?, 1)    f29(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f41(1, s, c, 0, r, r, r, h, i, j, k, l, m, n, o, p, q) [ 0 >= a /\ r >= c + 1000 ]
		(?, 1)    f21(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f29(0, b, r, d, e, f, g, 0, r, r, k, l, m, n, o, p, q) [ a >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f21(1, b, c, d, e, f, g, h, i, j, k, r, k, n, o, p, q) [ 0 >= k ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f21(1, b, c, d, e, f, g, h, i, j, s, r, 0, 1, s, s, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q) -> f41(1, b, c, d, e, f, g, h, i, j, s, r, 0, 1, s, s, s) [ 0 >= s /\ k >= 1 ]
	start location:	f0
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [a, c, k].
We thus obtain the following problem:
2:	T:
		(1, 1)    f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(?, 1)    f21(a, c, k) -> f29(0, r, k) [ a >= 1 ]
		(?, 1)    f29(a, c, k) -> f41(1, c, k) [ 0 >= a /\ r >= c + 1000 ]
		(?, 1)    f29(a, c, k) -> f41(a, c, k) [ 0 >= a /\ c + 999 >= r ]
		(?, 1)    f29(a, c, k) -> f41(a, c, k) [ a >= 1 ]
		(?, 1)    f43(a, c, k) -> f46(a, c, k)
		(?, 1)    f41(a, c, k) -> f41(a, c, k)
		(?, 1)    f21(a, c, k) -> f29(a, c, k) [ 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, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(?, 1)    f21(a, c, k) -> f29(0, r, k) [ a >= 1 ]
		(?, 1)    f29(a, c, k) -> f41(1, c, k) [ 0 >= a /\ r >= c + 1000 ]
		(?, 1)    f29(a, c, k) -> f41(a, c, k) [ 0 >= a /\ c + 999 >= r ]
		(?, 1)    f29(a, c, k) -> f41(a, c, k) [ a >= 1 ]
		(?, 1)    f41(a, c, k) -> f41(a, c, k)
		(?, 1)    f21(a, c, k) -> f29(a, c, k) [ 0 >= a ]
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transitions from problem 3:
	f29(a, c, k) -> f41(a, c, k) [ a >= 1 ]
	f21(a, c, k) -> f29(a, c, k) [ 0 >= a ]
We thus obtain the following problem:
4:	T:
		(?, 1)    f29(a, c, k) -> f41(a, c, k) [ 0 >= a /\ c + 999 >= r ]
		(?, 1)    f29(a, c, k) -> f41(1, c, k) [ 0 >= a /\ r >= c + 1000 ]
		(?, 1)    f21(a, c, k) -> f29(0, r, k) [ a >= 1 ]
		(?, 1)    f41(a, c, k) -> f41(a, c, k)
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ]
	start location:	f0
	leaf cost:	1

Repeatedly propagating knowledge in problem 4 produces the following problem:
5:	T:
		(2, 1)    f29(a, c, k) -> f41(a, c, k) [ 0 >= a /\ c + 999 >= r ]
		(2, 1)    f29(a, c, k) -> f41(1, c, k) [ 0 >= a /\ r >= c + 1000 ]
		(2, 1)    f21(a, c, k) -> f29(0, r, k) [ a >= 1 ]
		(?, 1)    f41(a, c, k) -> f41(a, c, k)
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ]
	start location:	f0
	leaf cost:	1

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f21: -X_1 + 1 >= 0 /\ X_1 - 1 >= 0
  For symbol f29: -X_1 >= 0 /\ X_1 >= 0
  For symbol f41: -X_1 + 1 >= 0 /\ X_1 >= 0


This yielded the following problem:
6:	T:
		(1, 1)    f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(?, 1)    f41(a, c, k) -> f41(a, c, k) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f21(a, c, k) -> f29(0, r, k) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f29(a, c, k) -> f41(1, c, k) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ r >= c + 1000 ]
		(2, 1)    f29(a, c, k) -> f41(a, c, k) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ c + 999 >= r ]
	start location:	f0
	leaf cost:	1

By chaining the transition f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 ] with all transitions in problem 6, the following new transition is obtained:
	f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
We thus obtain the following problem:
7:	T:
		(1, 2)    f0(a, c, k) -> f41(1, c, s) [ 0 >= s /\ k >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, s) [ s >= 1 /\ k >= 1 ]
		(1, 1)    f0(a, c, k) -> f21(1, c, k) [ 0 >= k ]
		(?, 1)    f41(a, c, k) -> f41(a, c, k) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f21(a, c, k) -> f29(0, r, k) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f29(a, c, k) -> f41(1, c, k) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ r >= c + 1000 ]
		(2, 1)    f29(a, c, k) -> f41(a, c, k) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ c + 999 >= r ]
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 0.380 sec (SMT: 0.359 sec)