YES(?, 8)

Initial complexity problem:
1:	T:
		(1, 1)    f0(a, b) -> f4(0, b)
		(?, 1)    f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
		(?, 1)    f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
		(?, 1)    f10(a, b) -> f18(a, b) [ b >= 2 /\ 0 >= c + 1 ]
		(?, 1)    f10(a, b) -> f18(a, b) [ b >= 2 ]
		(?, 1)    f4(a, b) -> f10(a, 0) [ a >= 2 ]
	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) -> f4(0, b)
		(?, 1)    f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
		(?, 1)    f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
		(?, 1)    f4(a, b) -> f10(a, 0) [ a >= 2 ]
	start location:	f0
	leaf cost:	2

A polynomial rank function with
	Pol(f0) = 1
	Pol(f4) = 1
	Pol(f10) = 0
orients all transitions weakly and the transition
	f4(a, b) -> f10(a, 0) [ a >= 2 ]
strictly and produces the following problem:
3:	T:
		(1, 1)    f0(a, b) -> f4(0, b)
		(?, 1)    f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
		(?, 1)    f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
		(1, 1)    f4(a, b) -> f10(a, 0) [ a >= 2 ]
	start location:	f0
	leaf cost:	2

A polynomial rank function with
	Pol(f0) = 2
	Pol(f4) = 2
	Pol(f10) = -V_2 + 2
orients all transitions weakly and the transition
	f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
strictly and produces the following problem:
4:	T:
		(1, 1)    f0(a, b) -> f4(0, b)
		(?, 1)    f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
		(2, 1)    f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
		(1, 1)    f4(a, b) -> f10(a, 0) [ a >= 2 ]
	start location:	f0
	leaf cost:	2

A polynomial rank function with
	Pol(f0) = 2
	Pol(f4) = -V_1 + 2
	Pol(f10) = -V_1
orients all transitions weakly and the transition
	f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
strictly and produces the following problem:
5:	T:
		(1, 1)    f0(a, b) -> f4(0, b)
		(2, 1)    f4(a, b) -> f4(a + 1, b) [ 1 >= a ]
		(2, 1)    f10(a, b) -> f10(a, b + 1) [ 1 >= b ]
		(1, 1)    f4(a, b) -> f10(a, 0) [ a >= 2 ]
	start location:	f0
	leaf cost:	2

Complexity upper bound 8

Time: 0.168 sec (SMT: 0.159 sec)