YES(?, b + 1)

Initial complexity problem:
1:	T:
		(1, 1)    f0(a, b) -> f1(a, b) [ a >= 1 ]
		(?, 1)    f1(a, b) -> f1(a + 1, -a + b) [ b >= 1 ]
	start location:	f0
	leaf cost:	0

Applied AI with 'oct' on problem 1 to obtain the following invariants:
  For symbol f1: X_1 - 1 >= 0


This yielded the following problem:
2:	T:
		(?, 1)    f1(a, b) -> f1(a + 1, -a + b) [ a - 1 >= 0 /\ b >= 1 ]
		(1, 1)    f0(a, b) -> f1(a, b) [ a >= 1 ]
	start location:	f0
	leaf cost:	0

A polynomial rank function with
	Pol(f1) = V_2
	Pol(f0) = V_2
orients all transitions weakly and the transition
	f1(a, b) -> f1(a + 1, -a + b) [ a - 1 >= 0 /\ b >= 1 ]
strictly and produces the following problem:
3:	T:
		(b, 1)    f1(a, b) -> f1(a + 1, -a + b) [ a - 1 >= 0 /\ b >= 1 ]
		(1, 1)    f0(a, b) -> f1(a, b) [ a >= 1 ]
	start location:	f0
	leaf cost:	0

Complexity upper bound b + 1

Time: 0.216 sec (SMT: 0.206 sec)