YES(?, 4*a + b + 1)

Initial complexity problem:
1:	T:
		(?, 1)    eval1(a, b) -> eval2(a, b) [ a >= 1 ]
		(?, 1)    eval2(a, b) -> eval2(a, b - 1) [ a >= 1 /\ b >= 1 ]
		(?, 1)    eval2(a, b) -> eval1(a - 1, b) [ a >= 1 /\ 0 >= b ]
		(1, 1)    start(a, b) -> eval1(a, b)
	start location:	start
	leaf cost:	0

A polynomial rank function with
	Pol(eval1) = 2*V_1
	Pol(eval2) = 2*V_1 - 1
	Pol(start) = 2*V_1
orients all transitions weakly and the transitions
	eval2(a, b) -> eval1(a - 1, b) [ a >= 1 /\ 0 >= b ]
	eval1(a, b) -> eval2(a, b) [ a >= 1 ]
strictly and produces the following problem:
2:	T:
		(2*a, 1)    eval1(a, b) -> eval2(a, b) [ a >= 1 ]
		(?, 1)      eval2(a, b) -> eval2(a, b - 1) [ a >= 1 /\ b >= 1 ]
		(2*a, 1)    eval2(a, b) -> eval1(a - 1, b) [ a >= 1 /\ 0 >= b ]
		(1, 1)      start(a, b) -> eval1(a, b)
	start location:	start
	leaf cost:	0

A polynomial rank function with
	Pol(eval1) = V_2
	Pol(eval2) = V_2
	Pol(start) = V_2
orients all transitions weakly and the transition
	eval2(a, b) -> eval2(a, b - 1) [ a >= 1 /\ b >= 1 ]
strictly and produces the following problem:
3:	T:
		(2*a, 1)    eval1(a, b) -> eval2(a, b) [ a >= 1 ]
		(b, 1)      eval2(a, b) -> eval2(a, b - 1) [ a >= 1 /\ b >= 1 ]
		(2*a, 1)    eval2(a, b) -> eval1(a - 1, b) [ a >= 1 /\ 0 >= b ]
		(1, 1)      start(a, b) -> eval1(a, b)
	start location:	start
	leaf cost:	0

Complexity upper bound 4*a + b + 1

Time: 0.151 sec (SMT: 0.142 sec)