YES(?, 66)

Initial complexity problem:
1:	T:
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j) -> f5(k, 0, 0, d, e, f, g, h, i, j)
		(?, 1)    f5(a, b, c, d, e, f, g, h, i, j) -> f5(a, b + 1, c + 1, 1, e, f, g, h, i, j) [ 31 >= c ]
		(?, 1)    f5(a, b, c, d, e, f, g, h, i, j) -> f5(a, b, c + 1, 0, e, f, g, h, i, j) [ 31 >= c ]
		(?, 1)    f5(a, b, c, d, e, f, g, h, i, j) -> f28(a, b, c, d, b, b, k, l, l, l) [ c >= 32 ]
	start location:	f0
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [c].
We thus obtain the following problem:
2:	T:
		(?, 1)    f5(c) -> f28(c) [ c >= 32 ]
		(?, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(?, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(1, 1)    f0(c) -> f5(0)
	start location:	f0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 2 produces the following problem:
3:	T:
		(?, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(?, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(1, 1)    f0(c) -> f5(0)
	start location:	f0
	leaf cost:	1

A polynomial rank function with
	Pol(f5) = -V_1 + 32
	Pol(f0) = 32
orients all transitions weakly and the transition
	f5(c) -> f5(c + 1) [ 31 >= c ]
strictly and produces the following problem:
4:	T:
		(32, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(32, 1)    f5(c) -> f5(c + 1) [ 31 >= c ]
		(1, 1)     f0(c) -> f5(0)
	start location:	f0
	leaf cost:	1

Complexity upper bound 66

Time: 0.121 sec (SMT: 0.117 sec)