YES(?, 12)

Initial complexity problem:
1:	T:
		(?, 1)    f0(a, b, c, d) -> f4(a, e, c, d) [ a >= 10 ]
		(?, 1)    f0(a, b, c, d) -> f0(a + 1, b, a, d) [ 9 >= a ]
		(?, 1)    f1(a, b, c, d) -> f0(1, b, c, d) [ 9 >= e /\ a = 0 ]
		(?, 1)    f2(a, b, c, d) -> f0(2, b, c, 2) [ 9 >= a ]
		(1, 1)    f3(a, b, c, d) -> f0(0, b, c, d)
	start location:	f3
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem:
2:	T:
		(?, 1)    f0(a, b, c, d) -> f0(a + 1, b, a, d) [ 9 >= a ]
		(?, 1)    f1(a, b, c, d) -> f0(1, b, c, d) [ 9 >= e /\ a = 0 ]
		(?, 1)    f2(a, b, c, d) -> f0(2, b, c, 2) [ 9 >= a ]
		(1, 1)    f3(a, b, c, d) -> f0(0, b, c, d)
	start location:	f3
	leaf cost:	1

Testing for reachability in the complexity graph removes the following transitions from problem 2:
	f1(a, b, c, d) -> f0(1, b, c, d) [ 9 >= e /\ a = 0 ]
	f2(a, b, c, d) -> f0(2, b, c, 2) [ 9 >= a ]
We thus obtain the following problem:
3:	T:
		(?, 1)    f0(a, b, c, d) -> f0(a + 1, b, a, d) [ 9 >= a ]
		(1, 1)    f3(a, b, c, d) -> f0(0, b, c, d)
	start location:	f3
	leaf cost:	1

A polynomial rank function with
	Pol(f0) = -V_1 + 10
	Pol(f3) = 10
orients all transitions weakly and the transition
	f0(a, b, c, d) -> f0(a + 1, b, a, d) [ 9 >= a ]
strictly and produces the following problem:
4:	T:
		(10, 1)    f0(a, b, c, d) -> f0(a + 1, b, a, d) [ 9 >= a ]
		(1, 1)     f3(a, b, c, d) -> f0(0, b, c, d)
	start location:	f3
	leaf cost:	1

Complexity upper bound 12

Time: 0.117 sec (SMT: 0.112 sec)