MAYBE

Initial complexity problem:
1:	T:
		(?, 1)    f49(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t)
		(?, 1)    f51(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f54(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t)
		(?, 1)    f11(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, 0, 0, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) [ a >= b ]
		(?, 1)    f35(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, 0, 0, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) [ e >= 3 ]
		(?, 1)    f35(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, 0, 0, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) [ 1 >= e ]
		(?, 1)    f35(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, 0, 0, 2, g, g, h, i, j, k, l, m, n, o, p, q, r, s, t) [ e = 2 ]
		(?, 1)    f11(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f49(a, b, 0, 0, e, f, u, v, w, x, d, u, u, n, o, p, q, r, s, t) [ u >= 1 /\ b >= a + 1 ]
		(?, 1)    f11(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f35(a, b, c, d, y, f, u, v, w, x, d, u, u, u, p, 0, y, y, 0, t) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(?, 1)    f11(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f35(a, b, c, d, y, f, u, v, w, x, d, u, u, u, p, 0, y, y, 0, t) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(?, 1)    f11(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f11(a + 1, b, c, d, 1, f, u, v, w, x, d, u, u, u, p, p, 1, 1, 0, t) [ 0 >= u /\ b >= a + 1 ]
		(1, 1)    f0(a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t) -> f11(a, b, 0, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, 0)
	start location:	f0
	leaf cost:	0

Slicing away variables that do not contribute to conditions from problem 1 leaves variables [a, b, e].
We thus obtain the following problem:
2:	T:
		(1, 1)    f0(a, b, e) -> f11(a, b, e)
		(?, 1)    f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
		(?, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(?, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(?, 1)    f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
		(?, 1)    f35(a, b, e) -> f49(a, b, 2) [ e = 2 ]
		(?, 1)    f35(a, b, e) -> f49(a, b, e) [ 1 >= e ]
		(?, 1)    f35(a, b, e) -> f49(a, b, e) [ e >= 3 ]
		(?, 1)    f11(a, b, e) -> f49(a, b, e) [ a >= b ]
		(?, 1)    f51(a, b, e) -> f54(a, b, e)
		(?, 1)    f49(a, b, e) -> f49(a, b, e)
	start location:	f0
	leaf cost:	0

Repeatedly removing leaves of the complexity graph in problem 2 produces the following problem:
3:	T:
		(1, 1)    f0(a, b, e) -> f11(a, b, e)
		(?, 1)    f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
		(?, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(?, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(?, 1)    f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
		(?, 1)    f35(a, b, e) -> f49(a, b, 2) [ e = 2 ]
		(?, 1)    f35(a, b, e) -> f49(a, b, e) [ 1 >= e ]
		(?, 1)    f35(a, b, e) -> f49(a, b, e) [ e >= 3 ]
		(?, 1)    f11(a, b, e) -> f49(a, b, e) [ a >= b ]
		(?, 1)    f49(a, b, e) -> f49(a, b, e)
	start location:	f0
	leaf cost:	1

A polynomial rank function with
	Pol(f0) = 2
	Pol(f11) = 2
	Pol(f35) = 1
	Pol(f49) = 0
orients all transitions weakly and the transitions
	f35(a, b, e) -> f49(a, b, 2) [ e = 2 ]
	f35(a, b, e) -> f49(a, b, e) [ 1 >= e ]
	f35(a, b, e) -> f49(a, b, e) [ e >= 3 ]
	f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
	f11(a, b, e) -> f49(a, b, e) [ a >= b ]
	f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
	f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
strictly and produces the following problem:
4:	T:
		(1, 1)    f0(a, b, e) -> f11(a, b, e)
		(?, 1)    f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
		(2, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(2, 1)    f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(2, 1)    f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
		(2, 1)    f35(a, b, e) -> f49(a, b, 2) [ e = 2 ]
		(2, 1)    f35(a, b, e) -> f49(a, b, e) [ 1 >= e ]
		(2, 1)    f35(a, b, e) -> f49(a, b, e) [ e >= 3 ]
		(2, 1)    f11(a, b, e) -> f49(a, b, e) [ a >= b ]
		(?, 1)    f49(a, b, e) -> f49(a, b, e)
	start location:	f0
	leaf cost:	1

A polynomial rank function with
	Pol(f0) = -V_1 + V_2
	Pol(f11) = -V_1 + V_2
	Pol(f35) = -V_1 + V_2
	Pol(f49) = -V_1 + V_2
orients all transitions weakly and the transition
	f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
strictly and produces the following problem:
5:	T:
		(1, 1)        f0(a, b, e) -> f11(a, b, e)
		(a + b, 1)    f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
		(2, 1)        f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(2, 1)        f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(2, 1)        f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
		(2, 1)        f35(a, b, e) -> f49(a, b, 2) [ e = 2 ]
		(2, 1)        f35(a, b, e) -> f49(a, b, e) [ 1 >= e ]
		(2, 1)        f35(a, b, e) -> f49(a, b, e) [ e >= 3 ]
		(2, 1)        f11(a, b, e) -> f49(a, b, e) [ a >= b ]
		(?, 1)        f49(a, b, e) -> f49(a, b, e)
	start location:	f0
	leaf cost:	1

Applied AI with 'oct' on problem 5 to obtain the following invariants:
  For symbol f35: -X_1 + X_2 - 1 >= 0


This yielded the following problem:
6:	T:
		(?, 1)        f49(a, b, e) -> f49(a, b, e)
		(2, 1)        f11(a, b, e) -> f49(a, b, e) [ a >= b ]
		(2, 1)        f35(a, b, e) -> f49(a, b, e) [ -a + b - 1 >= 0 /\ e >= 3 ]
		(2, 1)        f35(a, b, e) -> f49(a, b, e) [ -a + b - 1 >= 0 /\ 1 >= e ]
		(2, 1)        f35(a, b, e) -> f49(a, b, 2) [ -a + b - 1 >= 0 /\ e = 2 ]
		(2, 1)        f11(a, b, e) -> f49(a, b, e) [ u >= 1 /\ b >= a + 1 ]
		(2, 1)        f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ y >= 2 ]
		(2, 1)        f11(a, b, e) -> f35(a, b, y) [ b >= a + 1 /\ 0 >= u /\ 0 >= y ]
		(a + b, 1)    f11(a, b, e) -> f11(a + 1, b, 1) [ 0 >= u /\ b >= a + 1 ]
		(1, 1)        f0(a, b, e) -> f11(a, b, e)
	start location:	f0
	leaf cost:	1

Complexity upper bound ?

Time: 0.473 sec (SMT: 0.442 sec)