MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f2(a, b, c, d, e, f, g, h, i, j, k) -> f2(a, c, c, l, m, n, g, h, i, j, k) [ a >= 1 ]
		(1, 1)    f300(a, b, c, d, e, f, g, h, i, j, k) -> f1(a, l, n, p, q, f, g, l, m, o, r) [ 0 >= g ]
		(1, 1)    f300(a, b, c, d, e, f, g, h, i, j, k) -> f2(a, n, n, p, q, r, g, l, m, o, k) [ a >= 1 /\ g >= 1 ]
		(1, 1)    f300(a, b, c, d, e, f, g, h, i, j, k) -> f1(a, n, n, p, q, r, g, l, m, o, s) [ 0 >= a /\ g >= 1 ]
	start location:	f300
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [a, g].
We thus obtain the following problem:
2:	T:
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= a /\ g >= 1 ]
		(1, 1)    f300(a, g) -> f2(a, g) [ a >= 1 /\ g >= 1 ]
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= g ]
		(?, 1)    f2(a, g) -> f2(a, g) [ a >= 1 ]
	start location:	f300
	leaf cost:	0

Applied AI with 'oct' on problem 2 to obtain the following invariants:
  For symbol f2: X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0


This yielded the following problem:
3:	T:
		(?, 1)    f2(a, g) -> f2(a, g) [ g - 1 >= 0 /\ a + g - 2 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= g ]
		(1, 1)    f300(a, g) -> f2(a, g) [ a >= 1 /\ g >= 1 ]
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= a /\ g >= 1 ]
	start location:	f300
	leaf cost:	0

By chaining the transition f300(a, g) -> f2(a, g) [ a >= 1 /\ g >= 1 ] with all transitions in problem 3, the following new transition is obtained:
	f300(a, g) -> f2(a, g) [ a >= 1 /\ g >= 1 /\ g - 1 >= 0 /\ a + g - 2 >= 0 /\ a - 1 >= 0 ]
We thus obtain the following problem:
4:	T:
		(1, 2)    f300(a, g) -> f2(a, g) [ a >= 1 /\ g >= 1 /\ g - 1 >= 0 /\ a + g - 2 >= 0 /\ a - 1 >= 0 ]
		(?, 1)    f2(a, g) -> f2(a, g) [ g - 1 >= 0 /\ a + g - 2 >= 0 /\ a - 1 >= 0 /\ a >= 1 ]
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= g ]
		(1, 1)    f300(a, g) -> f1(a, g) [ 0 >= a /\ g >= 1 ]
	start location:	f300
	leaf cost:	0

Complexity upper bound ?

Time: 0.171 sec (SMT: 0.157 sec)