YES(?, 2*a + 7)

Initial complexity problem:
1:	T:
		(?, 1)    start(a, b, c, d, e, f) -> stop(a, b, c, d, e, f) [ 0 >= a /\ b = c /\ d = a /\ e = f ]
		(?, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, 1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(?, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, -1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> stop(a, b, c, d, e, f) [ e + a >= 0 /\ a >= 1 /\ a >= e /\ b = a + 1 /\ d = a ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e + 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e - 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(1, 1)    start0(a, b, c, d, e, f) -> start(a, c, c, a, f, f)
	start location:	start0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(?, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, 1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(?, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, -1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e + 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e - 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(1, 1)    start0(a, b, c, d, e, f) -> start(a, c, c, a, f, f)
	start location:	start0
	leaf cost:	2

Repeatedly propagating knowledge in problem 2 produces the following problem:
3:	T:
		(1, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, 1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(1, 1)    start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, -1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e + 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(?, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e - 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(1, 1)    start0(a, b, c, d, e, f) -> start(a, c, c, a, f, f)
	start location:	start0
	leaf cost:	2

A polynomial rank function with
	Pol(start) = V_1 - 1
	Pol(lbl101) = V_1 - V_2 + 1
	Pol(start0) = V_1 - 1
orients all transitions weakly and the transitions
	lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e + 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
	lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e - 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
strictly and produces the following problem:
4:	T:
		(1, 1)        start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, 1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(1, 1)        start(a, b, c, d, e, f) -> lbl101(a, 2, c, d, -1, f) [ a >= 1 /\ b = c /\ d = a /\ e = f ]
		(a + 1, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e + 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(a + 1, 1)    lbl101(a, b, c, d, e, f) -> lbl101(a, b + 1, c, d, e - 1, f) [ a >= b /\ e + b >= 1 /\ a + 1 >= b /\ b >= 2 /\ b >= e + 1 /\ d = a ]
		(1, 1)        start0(a, b, c, d, e, f) -> start(a, c, c, a, f, f)
	start location:	start0
	leaf cost:	2

Complexity upper bound 2*a + 7

Time: 0.292 sec (SMT: 0.275 sec)