MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f26(a, b) -> f2(a, b)
		(?, 1)    f2(a, b) -> f2(a, b)
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f28(a, b) -> f30(a, b)
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ 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)    f26(a, b) -> f2(a, b)
		(?, 1)    f2(a, b) -> f2(a, b)
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ a >= 6 ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ 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:
	f26(a, b) -> f2(a, b)
	f2(a, b) -> f2(a, b)
	f12(a, b) -> f9(a + 1, b) [ a >= 6 ]
We thus obtain the following problem:
3:	T:
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ 0 >= c + 1 ]
		(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 f12: -X_1 + 5 >= 0


This yielded the following problem:
4:	T:
		(1, 1)    f0(a, b) -> f9(c, b)
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ 0 >= c + 1 ]
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ a >= 6 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ -a + 5 >= 0 /\ 5 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f12(a, c) [ 5 >= a /\ 0 >= c + 1 ] with all transitions in problem 4, the following new transition is obtained:
	f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
We thus obtain the following problem:
5:	T:
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(1, 1)    f0(a, b) -> f9(c, b)
		(?, 1)    f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ a >= 6 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ -a + 5 >= 0 /\ 5 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f12(a, c) [ 5 >= a /\ c >= 1 ] with all transitions in problem 5, the following new transition is obtained:
	f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
We thus obtain the following problem:
6:	T:
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(1, 1)    f0(a, b) -> f9(c, b)
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(?, 1)    f9(a, b) -> f20(a, b) [ a >= 6 ]
		(?, 1)    f12(a, b) -> f9(a + 1, b) [ -a + 5 >= 0 /\ 5 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
	start location:	f0
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transition from problem 6:
	f12(a, b) -> f9(a + 1, b) [ -a + 5 >= 0 /\ 5 >= a ]
We thus obtain the following problem:
7:	T:
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, b) [ a >= 6 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a, b) [ a >= 6 ] with all transitions in problem 7, the following new transition is obtained:
	f9(a, b) -> f20(a - 1, b) [ a >= 6 /\ a >= 3 ]
We thus obtain the following problem:
8:	T:
		(?, 2)    f9(a, b) -> f20(a - 1, b) [ a >= 6 /\ a >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 1, b) [ a >= 6 /\ a >= 3 ] with all transitions in problem 8, the following new transition is obtained:
	f9(a, b) -> f20(a - 2, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 ]
We thus obtain the following problem:
9:	T:
		(?, 3)    f9(a, b) -> f20(a - 2, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 2, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 ] with all transitions in problem 9, the following new transition is obtained:
	f9(a, b) -> f20(a - 3, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ]
We thus obtain the following problem:
10:	T:
		(?, 4)    f9(a, b) -> f20(a - 3, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 3, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 ] with all transitions in problem 10, the following new transition is obtained:
	f9(a, b) -> f20(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
We thus obtain the following problem:
11:	T:
		(?, 5)    f9(a, b) -> f20(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 ] with all transitions in problem 11, the following new transitions are obtained:
	f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
	f9(a, b) -> f20(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ]
We thus obtain the following problem:
12:	T:
		(?, 6)    f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 6)    f9(a, b) -> f20(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Repeatedly propagating knowledge in problem 12 produces the following problem:
13:	T:
		(?, 6)    f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(1, 6)    f9(a, b) -> f20(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 ] with all transitions in problem 13, the following new transitions are obtained:
	f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
	f9(a, b) -> f20(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ]
We thus obtain the following problem:
14:	T:
		(1, 7)    f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(1, 7)    f9(a, b) -> f20(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ]
		(?, 6)    f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 ] with all transitions in problem 14, the following new transitions are obtained:
	f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
	f9(a, b) -> f20(a - 7, b) [ a >= 6 /\ 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:
15:	T:
		(1, 8)    f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 8)    f9(a, b) -> f20(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 ]
		(1, 7)    f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)    f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 7, b) [ a >= 6 /\ 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 15, the following new transitions are obtained:
	f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
	f9(a, b) -> f20(a - 8, b) [ a >= 6 /\ 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:
16:	T:
		(1, 9)    f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 9)    f9(a, b) -> f20(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 ]
		(1, 8)    f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)    f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)    f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)    f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)    f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)    f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)    f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)    f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 ] with all transitions in problem 16, the following new transitions are obtained:
	f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
	f9(a, b) -> f20(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 ]
We thus obtain the following problem:
17:	T:
		(1, 10)    f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
		(1, 10)    f9(a, b) -> f20(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 ]
		(1, 9)     f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 8)     f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)     f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)     f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)     f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)     f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)     f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 ] with all transitions in problem 17, the following new transitions are obtained:
	f9(a, b) -> f9(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ 2 >= a - 9 ]
	f9(a, b) -> f20(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 ]
We thus obtain the following problem:
18:	T:
		(1, 11)    f9(a, b) -> f9(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ 2 >= a - 9 ]
		(1, 11)    f9(a, b) -> f20(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 ]
		(1, 10)    f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
		(1, 9)     f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 8)     f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)     f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)     f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)     f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)     f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)     f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 ] with all transitions in problem 18, the following new transitions are obtained:
	f9(a, b) -> f9(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ 2 >= a - 10 ]
	f9(a, b) -> f20(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 ]
We thus obtain the following problem:
19:	T:
		(1, 12)    f9(a, b) -> f9(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ 2 >= a - 10 ]
		(1, 12)    f9(a, b) -> f20(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 ]
		(1, 11)    f9(a, b) -> f9(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ 2 >= a - 9 ]
		(1, 10)    f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
		(1, 9)     f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 8)     f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)     f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)     f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)     f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)     f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)     f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 ] with all transitions in problem 19, the following new transitions are obtained:
	f9(a, b) -> f9(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ 2 >= a - 11 ]
	f9(a, b) -> f20(a - 12, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 ]
We thus obtain the following problem:
20:	T:
		(1, 13)    f9(a, b) -> f9(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ 2 >= a - 11 ]
		(1, 13)    f9(a, b) -> f20(a - 12, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 ]
		(1, 12)    f9(a, b) -> f9(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ 2 >= a - 10 ]
		(1, 11)    f9(a, b) -> f9(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ 2 >= a - 9 ]
		(1, 10)    f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
		(1, 9)     f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 8)     f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)     f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)     f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)     f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)     f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)     f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

By chaining the transition f9(a, b) -> f20(a - 12, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 ] with all transitions in problem 20, the following new transitions are obtained:
	f9(a, b) -> f9(a - 12, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 /\ 2 >= a - 12 ]
	f9(a, b) -> f20(a - 13, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 /\ a - 12 >= 3 ]
We thus obtain the following problem:
21:	T:
		(1, 14)    f9(a, b) -> f9(a - 12, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 /\ 2 >= a - 12 ]
		(1, 14)    f9(a, b) -> f20(a - 13, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ a - 11 >= 3 /\ a - 12 >= 3 ]
		(1, 13)    f9(a, b) -> f9(a - 11, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ a - 10 >= 3 /\ 2 >= a - 11 ]
		(1, 12)    f9(a, b) -> f9(a - 10, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ a - 9 >= 3 /\ 2 >= a - 10 ]
		(1, 11)    f9(a, b) -> f9(a - 9, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ a - 8 >= 3 /\ 2 >= a - 9 ]
		(1, 10)    f9(a, b) -> f9(a - 8, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ a - 7 >= 3 /\ 2 >= a - 8 ]
		(1, 9)     f9(a, b) -> f9(a - 7, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ a - 6 >= 3 /\ 2 >= a - 7 ]
		(1, 8)     f9(a, b) -> f9(a - 6, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ a - 5 >= 3 /\ 2 >= a - 6 ]
		(1, 7)     f9(a, b) -> f9(a - 5, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ a - 4 >= 3 /\ 2 >= a - 5 ]
		(?, 6)     f9(a, b) -> f9(a - 4, b) [ a >= 6 /\ a >= 3 /\ a - 1 >= 3 /\ a - 2 >= 3 /\ a - 3 >= 3 /\ 2 >= a - 4 ]
		(?, 1)     f20(a, b) -> f9(a, b) [ 2 >= a ]
		(?, 1)     f20(a, b) -> f20(a - 1, b) [ a >= 3 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ c >= 1 /\ -a + 5 >= 0 ]
		(?, 2)     f9(a, b) -> f9(a + 1, c) [ 5 >= a /\ 0 >= c + 1 /\ -a + 5 >= 0 ]
		(?, 1)     f9(a, b) -> f20(a, 0) [ 5 >= a ]
		(1, 1)     f0(a, b) -> f9(c, b)
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 3.381 sec (SMT: 3.187 sec)