YES(?, a + b + 3)

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

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(?, 1)    eval1(a, b, c) -> eval2(a, b, c) [ a >= b + 1 /\ c = a ]
		(?, 1)    eval2(a, b, c) -> eval2(a - 1, b, c - 1) [ a >= b + 1 ]
		(1, 1)    start(a, b, c) -> eval1(a, b, c)
	start location:	start
	leaf cost:	1

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(1, 1)    eval1(a, b, c) -> eval2(a, b, c) [ a >= b + 1 /\ c = a ]
		(?, 1)    eval2(a, b, c) -> eval2(a - 1, b, c - 1) [ a >= b + 1 ]
		(1, 1)    start(a, b, c) -> eval1(a, b, c)
	start location:	start
	leaf cost:	1

A polynomial rank function with
	Pol(eval1) = V_1 - V_2
	Pol(eval2) = V_1 - V_2
	Pol(start) = V_1 - V_2
orients all transitions weakly and the transition
	eval2(a, b, c) -> eval2(a - 1, b, c - 1) [ a >= b + 1 ]
strictly and produces the following problem:
4:	T:
		(1, 1)        eval1(a, b, c) -> eval2(a, b, c) [ a >= b + 1 /\ c = a ]
		(a + b, 1)    eval2(a, b, c) -> eval2(a - 1, b, c - 1) [ a >= b + 1 ]
		(1, 1)        start(a, b, c) -> eval1(a, b, c)
	start location:	start
	leaf cost:	1

Complexity upper bound a + b + 3

Time: 0.084 sec (SMT: 0.080 sec)