MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f20(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f28(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) [ 0 >= a ]
		(?, 1)    f40(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f40(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o)
		(?, 1)    f42(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f45(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o)
		(?, 1)    f28(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f40(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) [ a >= 1 ]
		(?, 1)    f28(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f40(a, b, 0, p, p, p, g, h, i, j, k, l, m, n, o) [ 0 >= a /\ b + 999 >= p ]
		(?, 1)    f28(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f40(1, b, 0, p, p, p, g, h, i, j, k, l, m, n, o) [ 0 >= a /\ p >= b + 1000 ]
		(?, 1)    f20(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f28(0, p, c, d, e, f, 0, p, p, j, k, l, m, n, o) [ a >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f20(1, b, c, d, e, f, g, h, i, p, k, l, m, n, o) [ 0 >= p ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f20(1, b, c, d, e, f, g, h, i, 0, 1, p, p, p, p) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o) -> f40(1, b, c, d, e, f, g, h, i, 0, 1, p, p, p, p) [ 0 >= p /\ q >= 1 ]
	start location:	f0
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [a, b].
We thus obtain the following problem:
2:	T:
		(1, 1)    f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(?, 1)    f20(a, b) -> f28(0, p) [ a >= 1 ]
		(?, 1)    f28(a, b) -> f40(1, b) [ 0 >= a /\ p >= b + 1000 ]
		(?, 1)    f28(a, b) -> f40(a, b) [ 0 >= a /\ b + 999 >= p ]
		(?, 1)    f28(a, b) -> f40(a, b) [ a >= 1 ]
		(?, 1)    f42(a, b) -> f45(a, b)
		(?, 1)    f40(a, b) -> f40(a, b)
		(?, 1)    f20(a, b) -> f28(a, b) [ 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, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(?, 1)    f20(a, b) -> f28(0, p) [ a >= 1 ]
		(?, 1)    f28(a, b) -> f40(1, b) [ 0 >= a /\ p >= b + 1000 ]
		(?, 1)    f28(a, b) -> f40(a, b) [ 0 >= a /\ b + 999 >= p ]
		(?, 1)    f28(a, b) -> f40(a, b) [ a >= 1 ]
		(?, 1)    f40(a, b) -> f40(a, b)
		(?, 1)    f20(a, b) -> f28(a, b) [ 0 >= a ]
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transitions from problem 3:
	f28(a, b) -> f40(a, b) [ a >= 1 ]
	f20(a, b) -> f28(a, b) [ 0 >= a ]
We thus obtain the following problem:
4:	T:
		(?, 1)    f28(a, b) -> f40(a, b) [ 0 >= a /\ b + 999 >= p ]
		(?, 1)    f28(a, b) -> f40(1, b) [ 0 >= a /\ p >= b + 1000 ]
		(?, 1)    f20(a, b) -> f28(0, p) [ a >= 1 ]
		(?, 1)    f40(a, b) -> f40(a, b)
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ]
	start location:	f0
	leaf cost:	1

Repeatedly propagating knowledge in problem 4 produces the following problem:
5:	T:
		(2, 1)    f28(a, b) -> f40(a, b) [ 0 >= a /\ b + 999 >= p ]
		(2, 1)    f28(a, b) -> f40(1, b) [ 0 >= a /\ p >= b + 1000 ]
		(2, 1)    f20(a, b) -> f28(0, p) [ a >= 1 ]
		(?, 1)    f40(a, b) -> f40(a, b)
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ]
	start location:	f0
	leaf cost:	1

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f20: -X_1 + 1 >= 0 /\ X_1 - 1 >= 0
  For symbol f28: -X_1 >= 0 /\ X_1 >= 0
  For symbol f40: -X_1 + 1 >= 0 /\ X_1 >= 0


This yielded the following problem:
6:	T:
		(1, 1)    f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(?, 1)    f40(a, b) -> f40(a, b) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f20(a, b) -> f28(0, p) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f28(a, b) -> f40(1, b) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ p >= b + 1000 ]
		(2, 1)    f28(a, b) -> f40(a, b) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ b + 999 >= p ]
	start location:	f0
	leaf cost:	1

By chaining the transition f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 ] with all transitions in problem 6, the following new transition is obtained:
	f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
We thus obtain the following problem:
7:	T:
		(1, 2)    f0(a, b) -> f40(1, b) [ 0 >= p /\ q >= 1 /\ 0 >= 0 /\ 1 >= 0 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ p >= 1 /\ q >= 1 ]
		(1, 1)    f0(a, b) -> f20(1, b) [ 0 >= p ]
		(?, 1)    f40(a, b) -> f40(a, b) [ -a + 1 >= 0 /\ a >= 0 ]
		(2, 1)    f20(a, b) -> f28(0, p) [ -a + 1 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(2, 1)    f28(a, b) -> f40(1, b) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ p >= b + 1000 ]
		(2, 1)    f28(a, b) -> f40(a, b) [ -a >= 0 /\ a >= 0 /\ 0 >= a /\ b + 999 >= p ]
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 0.355 sec (SMT: 0.335 sec)