MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f9(a, b) -> f10(a, b) [ 5 >= a ]
		(?, 1)    f25(a, b) -> f2(a, b)
		(?, 1)    f2(a, b) -> f2(a, b)
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f27(a, b) -> f29(a, b)
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ a >= 3 ]
		(?, 1)    f19(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(?, 1)    f9(a, b) -> f10(a, b) [ 5 >= a ]
		(?, 1)    f25(a, b) -> f2(a, b)
		(?, 1)    f2(a, b) -> f2(a, b)
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ a >= 3 ]
		(?, 1)    f19(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f25(a, b) -> f2(a, b)
	f2(a, b) -> f2(a, b)
We thus obtain the following problem:
3:	T:
		(?, 1)    f19(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f10(a, b) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f19: -X_2 >= 0 /\ X_2 >= 0


This yielded the following problem:
4:	T:
		(1, 1)    f0(a, b) -> f9(c, b)
		(?, 1)    f9(a, b) -> f10(a, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f19(a, b) -> f9(a, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a ]
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f10(a, b) [ 5 >= a ] with all transitions in problem 4, the following new transition is obtained:
	f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
We thus obtain the following problem:
5:	T:
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f19(a, b) -> f9(a, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a ]
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transition from problem 5:
	f10(a, b) -> f9(a + 1, b) [ 5 >= a ]
We thus obtain the following problem:
6:	T:
		(?, 1)    f19(a, b) -> f9(a, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f19(a, b) -> f9(a, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a ] with all transitions in problem 6, the following new transition is obtained:
	f19(a, b) -> f9(a + 1, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a ]
We thus obtain the following problem:
7:	T:
		(?, 3)    f19(a, b) -> f9(a + 1, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f19(a, b) -> f9(a + 1, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a ] with all transitions in problem 7, the following new transition is obtained:
	f19(a, b) -> f9(a + 2, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 ]
We thus obtain the following problem:
8:	T:
		(?, 5)    f19(a, b) -> f9(a + 2, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f19(a, b) -> f9(a + 2, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 ] with all transitions in problem 8, the following new transition is obtained:
	f19(a, b) -> f9(a + 3, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 ]
We thus obtain the following problem:
9:	T:
		(?, 7)    f19(a, b) -> f9(a + 3, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f19(a, b) -> f9(a + 3, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 ] with all transitions in problem 9, the following new transition is obtained:
	f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
We thus obtain the following problem:
10:	T:
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f19(a, 0) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a, 0) [ a >= 6 ] with all transitions in problem 10, the following new transition is obtained:
	f9(a, b) -> f19(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 ]
We thus obtain the following problem:
11:	T:
		(?, 2)    f9(a, b) -> f19(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 ] with all transitions in problem 11, the following new transition is obtained:
	f9(a, b) -> f19(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 ]
We thus obtain the following problem:
12:	T:
		(?, 3)    f9(a, b) -> f19(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 ] with all transitions in problem 12, the following new transition is obtained:
	f9(a, b) -> f19(a - 3, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ]
We thus obtain the following problem:
13:	T:
		(?, 4)    f9(a, b) -> f19(a - 3, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 3, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ] with all transitions in problem 13, the following new transition is obtained:
	f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
We thus obtain the following problem:
14:	T:
		(?, 5)    f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f10(a, c) [ a >= 6 /\ c >= 1 ] with all transitions in problem 14, the following new transition is obtained:
	f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
We thus obtain the following problem:
15:	T:
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 5)    f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f10(a, c) [ a >= 6 /\ 0 >= c + 1 ] with all transitions in problem 15, the following new transition is obtained:
	f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
We thus obtain the following problem:
16:	T:
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 5)    f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 1)    f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transition from problem 16:
	f10(a, b) -> f9(a + 1, b) [ a >= 6 ]
We thus obtain the following problem:
17:	T:
		(?, 9)    f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)    f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 5)    f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 4, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ] with all transitions in problem 17, the following new transitions are obtained:
	f9(a, b) -> f9(a, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 /\ 5 >= a - 1 ]
	f9(a, b) -> f19(a - 5, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ]
We thus obtain the following problem:
18:	T:
		(?, 14)    f9(a, b) -> f9(a, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 /\ 5 >= a - 1 ]
		(?, 6)     f9(a, b) -> f19(a - 5, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ]
		(?, 9)     f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)     f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 5, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ] with all transitions in problem 18, the following new transitions are obtained:
	f9(a, b) -> f9(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 ]
	f9(a, b) -> f19(a - 6, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ]
We thus obtain the following problem:
19:	T:
		(?, 15)    f9(a, b) -> f9(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 ]
		(?, 7)     f9(a, b) -> f19(a - 6, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ]
		(?, 14)    f9(a, b) -> f9(a, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 /\ 5 >= a - 1 ]
		(?, 9)     f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)     f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 6, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ] with all transitions in problem 19, the following new transitions are obtained:
	f9(a, b) -> f9(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 /\ 5 >= a - 6 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 ]
	f9(a, b) -> f19(a - 7, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 ]
We thus obtain the following problem:
20:	T:
		(?, 16)    f9(a, b) -> f9(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 /\ 5 >= a - 6 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 ]
		(?, 8)     f9(a, b) -> f19(a - 7, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 ]
		(?, 15)    f9(a, b) -> f9(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 ]
		(?, 14)    f9(a, b) -> f9(a, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 /\ 5 >= a - 1 ]
		(?, 9)     f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)     f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f19(a - 7, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 ] with all transitions in problem 20, the following new transitions are obtained:
	f9(a, b) -> f9(a - 3, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 /\ 5 >= a - 7 /\ 5 >= a - 6 /\ 5 >= a - 5 /\ 5 >= a - 4 ]
	f9(a, b) -> f19(a - 8, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 ]
We thus obtain the following problem:
21:	T:
		(?, 17)    f9(a, b) -> f9(a - 3, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 /\ 5 >= a - 7 /\ 5 >= a - 6 /\ 5 >= a - 5 /\ 5 >= a - 4 ]
		(?, 9)     f9(a, b) -> f19(a - 8, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 ]
		(?, 16)    f9(a, b) -> f9(a - 2, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 /\ 5 >= a - 6 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 ]
		(?, 15)    f9(a, b) -> f9(a - 1, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 /\ 5 >= a - 5 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 ]
		(?, 14)    f9(a, b) -> f9(a, 0) [ a >= 6 /\ 0 >= 0 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 /\ 5 >= a - 4 /\ 5 >= a - 3 /\ 5 >= a - 2 /\ 5 >= a - 1 ]
		(?, 9)     f19(a, b) -> f9(a + 4, b) [ -b >= 0 /\ b >= 0 /\ 2 >= a /\ 5 >= a /\ 5 >= a + 1 /\ 5 >= a + 2 /\ 5 >= a + 3 ]
		(?, 1)     f19(a, b) -> f19(a - 1, b) [ -b >= 0 /\ b >= 0 /\ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ 0 >= c + 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ a >= 6 /\ c >= 1 ]
		(?, 2)     f9(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 2.832 sec (SMT: 2.654 sec)