MAYBE

Initial complexity problem:
1:	T:
		(1, 1)    f0(a, b, c, d) -> f6(b, b, d, d)
		(?, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ 0 >= a + 1 ]
		(?, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ a >= 1 ]
		(?, 1)    f6(a, b, c, d) -> f14(0, b, c, d) [ d >= b + 1 /\ a = 0 ]
		(?, 1)    f6(a, b, c, d) -> f14(0, b, c, d) [ b >= d + 1 /\ a = 0 ]
		(?, 1)    f6(a, b, c, d) -> f14(0, b, c, b) [ a = 0 /\ b = d ]
	start location:	f0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(1, 1)    f0(a, b, c, d) -> f6(b, b, d, d)
		(?, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ 0 >= a + 1 ]
		(?, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ a >= 1 ]
	start location:	f0
	leaf cost:	3

A polynomial rank function with
	Pol(f0) = 2*V_2 + 1
	Pol(f6) = 2*V_1 + 1
orients all transitions weakly and the transition
	f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ a >= 1 ]
strictly and produces the following problem:
3:	T:
		(1, 1)          f0(a, b, c, d) -> f6(b, b, d, d)
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ 0 >= a + 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ a >= 1 ]
	start location:	f0
	leaf cost:	3

Applied AI with 'oct' on problem 3 to obtain the following invariants:
  For symbol f6: -X_3 + X_4 >= 0 /\ -X_1 + X_2 >= 0


This yielded the following problem:
4:	T:
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
		(1, 1)          f0(a, b, c, d) -> f6(b, b, d, d)
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b, b, d, d) with all transitions in problem 4, the following new transitions are obtained:
	f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ 0 >= b + 1 ]
	f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
We thus obtain the following problem:
5:	T:
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ 0 >= b + 1 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ 0 >= b + 1 ] with all transitions in problem 5, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 2, b, d - 2, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b ]
We thus obtain the following problem:
6:	T:
		(1, 3)          f0(a, b, c, d) -> f6(b - 2, b, d - 2, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 2, b, d - 2, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b ] with all transitions in problem 6, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 3, b, d - 3, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 ]
We thus obtain the following problem:
7:	T:
		(1, 4)          f0(a, b, c, d) -> f6(b - 3, b, d - 3, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 3, b, d - 3, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 ] with all transitions in problem 7, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 4, b, d - 4, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 ]
We thus obtain the following problem:
8:	T:
		(1, 5)          f0(a, b, c, d) -> f6(b - 4, b, d - 4, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 4, b, d - 4, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 ] with all transitions in problem 8, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 5, b, d - 5, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 ]
We thus obtain the following problem:
9:	T:
		(1, 6)          f0(a, b, c, d) -> f6(b - 5, b, d - 5, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 5, b, d - 5, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 ] with all transitions in problem 9, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 6, b, d - 6, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 ]
We thus obtain the following problem:
10:	T:
		(1, 7)          f0(a, b, c, d) -> f6(b - 6, b, d - 6, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 6, b, d - 6, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 ] with all transitions in problem 10, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 7, b, d - 7, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 ]
We thus obtain the following problem:
11:	T:
		(1, 8)          f0(a, b, c, d) -> f6(b - 7, b, d - 7, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 7, b, d - 7, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 ] with all transitions in problem 11, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 8, b, d - 8, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 ]
We thus obtain the following problem:
12:	T:
		(1, 9)          f0(a, b, c, d) -> f6(b - 8, b, d - 8, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 8, b, d - 8, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 ] with all transitions in problem 12, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 9, b, d - 9, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 ]
We thus obtain the following problem:
13:	T:
		(1, 10)         f0(a, b, c, d) -> f6(b - 9, b, d - 9, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 9, b, d - 9, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 ] with all transitions in problem 13, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 10, b, d - 10, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 ]
We thus obtain the following problem:
14:	T:
		(1, 11)         f0(a, b, c, d) -> f6(b - 10, b, d - 10, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 10, b, d - 10, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 ] with all transitions in problem 14, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 11, b, d - 11, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 ]
We thus obtain the following problem:
15:	T:
		(1, 12)         f0(a, b, c, d) -> f6(b - 11, b, d - 11, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 11, b, d - 11, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 ] with all transitions in problem 15, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 12, b, d - 12, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 ]
We thus obtain the following problem:
16:	T:
		(1, 13)         f0(a, b, c, d) -> f6(b - 12, b, d - 12, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 12, b, d - 12, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 ] with all transitions in problem 16, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 13, b, d - 13, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 ]
We thus obtain the following problem:
17:	T:
		(1, 14)         f0(a, b, c, d) -> f6(b - 13, b, d - 13, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 13, b, d - 13, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 ] with all transitions in problem 17, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 14, b, d - 14, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 /\ 13 >= 0 /\ 0 >= b - 12 ]
We thus obtain the following problem:
18:	T:
		(1, 15)         f0(a, b, c, d) -> f6(b - 14, b, d - 14, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 /\ 13 >= 0 /\ 0 >= b - 12 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

By chaining the transition f0(a, b, c, d) -> f6(b - 14, b, d - 14, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 /\ 13 >= 0 /\ 0 >= b - 12 ] with all transitions in problem 18, the following new transition is obtained:
	f0(a, b, c, d) -> f6(b - 15, b, d - 15, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 /\ 13 >= 0 /\ 0 >= b - 12 /\ 14 >= 0 /\ 0 >= b - 13 ]
We thus obtain the following problem:
19:	T:
		(1, 16)         f0(a, b, c, d) -> f6(b - 15, b, d - 15, d) [ 0 >= 0 /\ 0 >= b + 1 /\ 1 >= 0 /\ 0 >= b /\ 2 >= 0 /\ 0 >= b - 1 /\ 3 >= 0 /\ 0 >= b - 2 /\ 4 >= 0 /\ 0 >= b - 3 /\ 5 >= 0 /\ 0 >= b - 4 /\ 6 >= 0 /\ 0 >= b - 5 /\ 7 >= 0 /\ 0 >= b - 6 /\ 8 >= 0 /\ 0 >= b - 7 /\ 9 >= 0 /\ 0 >= b - 8 /\ 10 >= 0 /\ 0 >= b - 9 /\ 11 >= 0 /\ 0 >= b - 10 /\ 12 >= 0 /\ 0 >= b - 11 /\ 13 >= 0 /\ 0 >= b - 12 /\ 14 >= 0 /\ 0 >= b - 13 ]
		(1, 2)          f0(a, b, c, d) -> f6(b - 1, b, d - 1, d) [ 0 >= 0 /\ b >= 1 ]
		(2*b + 1, 1)    f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ a >= 1 ]
		(?, 1)          f6(a, b, c, d) -> f6(a - 1, b, c - 1, d) [ -c + d >= 0 /\ -a + b >= 0 /\ 0 >= a + 1 ]
	start location:	f0
	leaf cost:	3

Complexity upper bound ?

Time: 1.724 sec (SMT: 1.614 sec)