YES(?, a + 3)

Initial complexity problem:
1:	T:
		(?, 1)    f2(a, b, c) -> f2(-b + a, b + 1, c) [ a >= 1 ]
		(1, 1)    f3(a, b, c) -> f2(a, b, c) [ b >= 1 ]
		(?, 1)    f2(a, b, c) -> f4(a, b, d) [ 0 >= a ]
		(1, 1)    f3(a, b, c) -> f4(a, b, d) [ 0 >= b ]
	start location:	f3
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(?, 1)    f2(a, b, c) -> f2(-b + a, b + 1, c) [ a >= 1 ]
		(1, 1)    f3(a, b, c) -> f2(a, b, c) [ b >= 1 ]
	start location:	f3
	leaf cost:	2

Applied AI with 'oct' on problem 2 to obtain the following invariants:
  For symbol f2: X_2 - 1 >= 0


This yielded the following problem:
3:	T:
		(1, 1)    f3(a, b, c) -> f2(a, b, c) [ b >= 1 ]
		(?, 1)    f2(a, b, c) -> f2(-b + a, b + 1, c) [ b - 1 >= 0 /\ a >= 1 ]
	start location:	f3
	leaf cost:	2

A polynomial rank function with
	Pol(f3) = V_1
	Pol(f2) = V_1
orients all transitions weakly and the transition
	f2(a, b, c) -> f2(-b + a, b + 1, c) [ b - 1 >= 0 /\ a >= 1 ]
strictly and produces the following problem:
4:	T:
		(1, 1)    f3(a, b, c) -> f2(a, b, c) [ b >= 1 ]
		(a, 1)    f2(a, b, c) -> f2(-b + a, b + 1, c) [ b - 1 >= 0 /\ a >= 1 ]
	start location:	f3
	leaf cost:	2

Complexity upper bound a + 3

Time: 0.240 sec (SMT: 0.228 sec)