2.34/2.42	MAYBE
2.34/2.42	
2.34/2.42	DP problem for innermost termination.
2.34/2.42	P =
2.34/2.42	  cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2#(false, x, y) -> ack#(x, y - 1) 
2.34/2.42	  cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
2.34/2.42	  cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
2.34/2.42	  ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
2.34/2.42	  ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f#(true, I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	  f#(true, I12) -> ack#(10, 10) 
2.34/2.42	R = 
2.34/2.42	  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
2.34/2.42	  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
2.34/2.42	  cond1(true, I4, I5) -> I5 + 1 
2.34/2.42	  ackNat(false, I6, I7) -> 0 
2.34/2.42	  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
2.34/2.42	  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	
2.34/2.42	This problem is converted using chaining, where edges between chained DPs are removed.
2.34/2.42	
2.34/2.42	DP problem for innermost termination.
2.34/2.42	P =
2.34/2.42	  cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2#(false, x, y) -> ack#(x, y - 1) 
2.34/2.42	  cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
2.34/2.42	  cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
2.34/2.42	  ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
2.34/2.42	  ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f#(true, I12) -> f_1#(I12) 
2.34/2.42	  f#(true, I12) -> ack#(10, 10) 
2.34/2.42	  f_1#(I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	  ack#(I10, I11) -> cond1#(I10 = 0, I10, I11) [I10 >= 0 && I11 >= 0]
2.34/2.42	  ack#(I10, I11) -> cond2#(I11 = 0, I10, I11) [I10 >= 0 && I11 >= 0, not(I10 = 0)]
2.34/2.42	  ack#(I10, I11) -> ack#(I10 - 1, I11 + 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), I11 = 0]
2.34/2.42	  ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	  ack#(I10, I11) -> ack#(I10 - 1, ack(I10, I11 - 1)) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	  ackNat#(true, I8, I9) -> cond2#(I9 = 0, I8, I9) [not(I8 = 0)]
2.34/2.42	  ackNat#(true, I8, I9) -> ack#(I8 - 1, I9 + 1) [not(I8 = 0), I9 = 0]
2.34/2.42	  ackNat#(true, I8, I9) -> ack#(I8, I9 - 1) [not(I8 = 0), not(I9 = 0)]
2.34/2.42	  ackNat#(true, I8, I9) -> ack#(I8 - 1, ack(I8, I9 - 1)) [not(I8 = 0), not(I9 = 0)]
2.34/2.42	  cond1#(false, I2, I3) -> ack#(I2 - 1, ack(I2, I3 - 1)) [not(I3 = 0)]
2.34/2.42	  cond1#(false, I2, I3) -> ack#(I2, I3 - 1) [not(I3 = 0)]
2.34/2.42	  cond1#(false, I2, I3) -> ack#(I2 - 1, I3 + 1) [I3 = 0]
2.34/2.42	R = 
2.34/2.42	  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
2.34/2.42	  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
2.34/2.42	  cond1(true, I4, I5) -> I5 + 1 
2.34/2.42	  ackNat(false, I6, I7) -> 0 
2.34/2.42	  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
2.34/2.42	  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	
2.34/2.42	The dependency graph for this problem is:
2.34/2.42	  0 -> 13, 12, 11, 10, 9, 5
2.34/2.42	  1 -> 13, 12, 11, 10, 9, 5
2.34/2.42	  2 -> 13, 12, 11, 10, 9, 5
2.34/2.42	  3 -> 
2.34/2.42	  4 -> 
2.34/2.42	  5 -> 
2.34/2.42	  6 -> 8
2.34/2.42	  7 -> 13, 12, 10, 9, 5
2.34/2.42	  8 -> 7, 6
2.34/2.42	  9 -> 
2.34/2.42	  10 -> 
2.34/2.42	  11 -> 13, 12, 5, 9, 10
2.34/2.42	  12 -> 13, 12, 5, 9, 10, 11
2.34/2.42	  13 -> 13, 5, 9, 10, 11, 12
2.34/2.42	  14 -> 
2.34/2.42	  15 -> 5, 9, 10, 12, 13
2.34/2.42	  16 -> 5, 9, 10, 11, 12, 13
2.34/2.42	  17 -> 5, 9, 10, 11, 12, 13
2.34/2.42	  18 -> 5, 9, 10, 11, 12, 13
2.34/2.42	  19 -> 5, 9, 10, 11, 12, 13
2.34/2.42	  20 -> 5, 9, 10, 12, 13
2.34/2.42	Where:
2.34/2.42	  0) cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
2.34/2.42	  1) cond2#(false, x, y) -> ack#(x, y - 1) 
2.34/2.42	  2) cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
2.34/2.42	  3) cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
2.34/2.42	  4) ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
2.34/2.42	  5) ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  6) f#(true, I12) -> f_1#(I12) 
2.34/2.42	  7) f#(true, I12) -> ack#(10, 10) 
2.34/2.42	  8) f_1#(I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	  9) ack#(I10, I11) -> cond1#(I10 = 0, I10, I11) [I10 >= 0 && I11 >= 0]
2.34/2.42	  10) ack#(I10, I11) -> cond2#(I11 = 0, I10, I11) [I10 >= 0 && I11 >= 0, not(I10 = 0)]
2.34/2.42	  11) ack#(I10, I11) -> ack#(I10 - 1, I11 + 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), I11 = 0]
2.34/2.42	  12) ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	  13) ack#(I10, I11) -> ack#(I10 - 1, ack(I10, I11 - 1)) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	  14) ackNat#(true, I8, I9) -> cond2#(I9 = 0, I8, I9) [not(I8 = 0)]
2.34/2.42	  15) ackNat#(true, I8, I9) -> ack#(I8 - 1, I9 + 1) [not(I8 = 0), I9 = 0]
2.34/2.42	  16) ackNat#(true, I8, I9) -> ack#(I8, I9 - 1) [not(I8 = 0), not(I9 = 0)]
2.34/2.42	  17) ackNat#(true, I8, I9) -> ack#(I8 - 1, ack(I8, I9 - 1)) [not(I8 = 0), not(I9 = 0)]
2.34/2.42	  18) cond1#(false, I2, I3) -> ack#(I2 - 1, ack(I2, I3 - 1)) [not(I3 = 0)]
2.34/2.42	  19) cond1#(false, I2, I3) -> ack#(I2, I3 - 1) [not(I3 = 0)]
2.34/2.42	  20) cond1#(false, I2, I3) -> ack#(I2 - 1, I3 + 1) [I3 = 0]
2.34/2.42	
2.34/2.42	We have the following SCCs.
2.34/2.42	  { 6, 8 }
2.34/2.42	  { 11, 12, 13 }
2.34/2.42	
2.34/2.42	DP problem for innermost termination.
2.34/2.42	P =
2.34/2.42	  ack#(I10, I11) -> ack#(I10 - 1, I11 + 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), I11 = 0]
2.34/2.42	  ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	  ack#(I10, I11) -> ack#(I10 - 1, ack(I10, I11 - 1)) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	R = 
2.34/2.42	  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
2.34/2.42	  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
2.34/2.42	  cond1(true, I4, I5) -> I5 + 1 
2.34/2.42	  ackNat(false, I6, I7) -> 0 
2.34/2.42	  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
2.34/2.42	  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	
2.34/2.42	We use the reverse value criterion with the projection function NU:
2.34/2.42	  NU[ack#(z1,z2)] = z1
2.34/2.42	
2.34/2.42	This gives the following inequalities:
2.34/2.42	  I10 >= 0 && I11 >= 0/\not(I10 = 0)/\I11 = 0  ==>  I10 > I10 - 1  with  I10 >= 0
2.34/2.42	  I10 >= 0 && I11 >= 0/\not(I10 = 0)/\not(I11 = 0)  ==>  I10 >= I10
2.34/2.42	  I10 >= 0 && I11 >= 0/\not(I10 = 0)/\not(I11 = 0)  ==>  I10 > I10 - 1  with  I10 >= 0
2.34/2.42	
2.34/2.42	We remove all the strictly oriented dependency pairs.
2.34/2.42	
2.34/2.42	DP problem for innermost termination.
2.34/2.42	P =
2.34/2.42	  ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
2.34/2.42	R = 
2.34/2.42	  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
2.34/2.42	  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
2.34/2.42	  cond1(true, I4, I5) -> I5 + 1 
2.34/2.42	  ackNat(false, I6, I7) -> 0 
2.34/2.42	  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
2.34/2.42	  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	
2.34/2.42	We use the reverse value criterion with the projection function NU:
2.34/2.42	  NU[ack#(z1,z2)] = z2
2.34/2.42	
2.34/2.42	This gives the following inequalities:
2.34/2.42	  I10 >= 0 && I11 >= 0/\not(I10 = 0)/\not(I11 = 0)  ==>  I11 > I11 - 1  with  I11 >= 0
2.34/2.42	
2.34/2.42	All dependency pairs are strictly oriented, so the entire dependency pair problem may be removed.
2.34/2.42	
2.34/2.42	DP problem for innermost termination.
2.34/2.42	P =
2.34/2.42	  f#(true, I12) -> f_1#(I12) 
2.34/2.42	  f_1#(I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	R = 
2.34/2.42	  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
2.34/2.42	  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
2.34/2.42	  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
2.34/2.42	  cond1(true, I4, I5) -> I5 + 1 
2.34/2.42	  ackNat(false, I6, I7) -> 0 
2.34/2.42	  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
2.34/2.42	  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
2.34/2.42	  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 
2.34/2.42	
2.34/5.39	EOF