MAYBE

Initial complexity problem:
1:	T:
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(1, 1)    f0(a, b) -> f2(1, c) [ 0 >= 4 /\ 0 >= 1 ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ 0 >= c /\ c >= 3 ]
		(1, 1)    f0(a, b) -> f2(1, c) [ 0 >= 4 /\ 0 >= 1 ]
		(1, 1)    f0(a, b) -> f2(3, c)
		(1, 1)    f0(a, b) -> f2(1, c) [ 0 >= 1 ]
		(1, 1)    f0(a, b) -> f2(3, c) [ 0 >= 3 ]
		(1, 1)    f0(a, b) -> f2(1, c) [ 0 >= 1 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ b >= 2 /\ 1 >= b /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ 0 >= b /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ 0 >= b /\ 1 >= b /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ 3 >= b /\ b >= 2 /\ 1 >= b /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ 0 >= 3 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ 3 >= b /\ 0 >= b /\ 1 >= b /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 0 >= 6*b + 4 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 0 >= 6*b + 4 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 6*b + 2 >= 0 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 0 >= 6*b + 4 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 0 >= 6*b + 4 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

Testing for unsatisfiable constraints removes the following transitions from problem 1:
	f0(a, b) -> f2(1, c) [ 0 >= 4 /\ 0 >= 1 ]
	f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ 0 >= c /\ c >= 3 ]
	f0(a, b) -> f2(1, c) [ 0 >= 4 /\ 0 >= 1 ]
	f0(a, b) -> f2(1, c) [ 0 >= 1 ]
	f0(a, b) -> f2(3, c) [ 0 >= 3 ]
	f0(a, b) -> f2(1, c) [ 0 >= 1 ]
	f2(a, b) -> f2(b, c) [ b >= 5 /\ b >= 2 /\ 1 >= b /\ a = 2*b ]
	f2(a, b) -> f2(b, c) [ b >= 5 /\ 0 >= b /\ b >= 3 /\ a = 2*b ]
	f2(a, b) -> f2(b, c) [ b >= 5 /\ 0 >= b /\ 1 >= b /\ a = 2*b ]
	f2(a, b) -> f2(b, c) [ 3 >= b /\ b >= 2 /\ 1 >= b /\ a = 2*b ]
	f2(a, b) -> f2(b, c) [ 0 >= 3 /\ b = 3 /\ a = 6 ]
	f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
	f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 0 >= 6*b + 4 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 0 >= 6*b + 4 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
	f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 6*b + 2 >= 0 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
	f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 0 >= 6*b + 4 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
We thus obtain the following problem:
2:	T:
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(1, 1)    f0(a, b) -> f2(3, c)
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ 3 >= b /\ 0 >= b /\ 1 >= b /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 0 >= 6*b + 4 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f2(a, b) -> f2(b, c) [ 3 >= b /\ 0 >= b /\ 1 >= b /\ a = 2*b ]
	f2(a, b) -> f2(6*b + 4, c) [ 0 >= 6*b + 1 /\ 0 >= 6*b + 4 /\ 0 >= 6*b + 3 /\ a = 2*b + 1 ]
We thus obtain the following problem:
3:	T:
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(1, 1)    f0(a, b) -> f2(3, c)
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
	start location:	f0
	leaf cost:	0

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f2: X_1 - 3 >= 0


This yielded the following problem:
4:	T:
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(1, 1)    f0(a, b) -> f2(3, c)
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f0(a, b) -> f2(3, c) with all transitions in problem 4, the following new transition is obtained:
	f0(a, b) -> f2(6*c + 4, c') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 ]
We thus obtain the following problem:
5:	T:
		(1, 2)    f0(a, b) -> f2(6*c + 4, c') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f0(a, b) -> f2(6*c + 4, c') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 ] with all transitions in problem 5, the following new transition is obtained:
	f0(a, b) -> f2(c', c'') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ]
We thus obtain the following problem:
6:	T:
		(1, 3)    f0(a, b) -> f2(c', c'') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f0(a, b) -> f2(c', c'') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ] with all transitions in problem 6, the following new transition is obtained:
	f0(a, b) -> f2(6*c'' + 4, c''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ]
We thus obtain the following problem:
7:	T:
		(1, 4)    f0(a, b) -> f2(6*c'' + 4, c''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f0(a, b) -> f2(6*c'' + 4, c''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ] with all transitions in problem 7, the following new transition is obtained:
	f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
We thus obtain the following problem:
8:	T:
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b = 3 /\ a = 6 ] with all transitions in problem 8, the following new transition is obtained:
	f2(a, b) -> f2(6*c + 4, c') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 ]
We thus obtain the following problem:
9:	T:
		(?, 2)    f2(a, b) -> f2(6*c + 4, c') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 ]
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f2(a, b) -> f2(6*c + 4, c') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 ] with all transitions in problem 9, the following new transition is obtained:
	f2(a, b) -> f2(c', c'') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ]
We thus obtain the following problem:
10:	T:
		(?, 3)    f2(a, b) -> f2(c', c'') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ]
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f2(a, b) -> f2(c', c'') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' ] with all transitions in problem 10, the following new transition is obtained:
	f2(a, b) -> f2(6*c'' + 4, c''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ]
We thus obtain the following problem:
11:	T:
		(?, 4)    f2(a, b) -> f2(6*c'' + 4, c''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ]
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f2(a, b) -> f2(6*c'' + 4, c''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 ] with all transitions in problem 11, the following new transition is obtained:
	f2(a, b) -> f2(c''', c'''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
We thus obtain the following problem:
12:	T:
		(?, 5)    f2(a, b) -> f2(c''', c'''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
		(?, 1)    f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ]
	start location:	f0
	leaf cost:	0

By chaining the transition f2(a, b) -> f2(6*b + 4, c) [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 ] with all transitions in problem 12, the following new transition is obtained:
	f2(a, b) -> f2(c, c') [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 /\ 6*b + 4 = 2*c ]
We thus obtain the following problem:
13:	T:
		(?, 2)    f2(a, b) -> f2(c, c') [ a - 3 >= 0 /\ 6*b >= 1 /\ 6*b + 2 >= 0 /\ 6*b + 1 >= 0 /\ a = 2*b + 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 /\ 6*b + 4 = 2*c ]
		(?, 5)    f2(a, b) -> f2(c''', c'''') [ a - 3 >= 0 /\ b = 3 /\ a = 6 /\ b - 3 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ b = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 5)    f0(a, b) -> f2(c''', c'''') [ 0 >= 0 /\ 6*c >= 1 /\ 6*c + 2 >= 0 /\ 6*c + 1 >= 0 /\ 3 = 2*c + 1 /\ c' >= 5 /\ c' >= 2 /\ c' >= 3 /\ 6*c + 4 = 2*c' /\ c' - 3 >= 0 /\ 6*c'' >= 1 /\ 6*c'' + 2 >= 0 /\ 6*c'' + 1 >= 0 /\ c' = 2*c'' + 1 /\ c''' >= 5 /\ c''' >= 2 /\ c''' >= 3 /\ 6*c'' + 4 = 2*c''' ]
		(1, 1)    f0(a, b) -> f2(c, d) [ c >= 1 /\ c >= 5 /\ c >= 2 /\ c >= 3 ]
		(?, 1)    f2(a, b) -> f2(b, c) [ a - 3 >= 0 /\ b >= 5 /\ b >= 2 /\ b >= 3 /\ a = 2*b ]
	start location:	f0
	leaf cost:	0

Complexity upper bound ?

Time: 2.307 sec (SMT: 2.168 sec)