Integer TRS Innermost pair #487100419

details

property	value
status	complete
benchmark	ackTen.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n138.star.cs.uiowa.edu
space	Mixed_ITRS_2014

run statistics

property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	2.49544 seconds
cpu usage	2.30625
user time	1.22227
system time	1.08398
max virtual memory	686232.0
max residence set size	8416.0

stage attributes

key	value
starexec-result	MAYBE

output

MAYBE

DP problem for innermost termination.
P =
  cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
  cond2#(false, x, y) -> ack#(x, y - 1) 
  cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
  cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
  ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
  ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
  f#(true, I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
  f#(true, I12) -> ack#(10, 10) 
R = 
  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
  cond1(true, I4, I5) -> I5 + 1 
  ackNat(false, I6, I7) -> 0 
  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 

This problem is converted using chaining, where edges between chained DPs are removed.

DP problem for innermost termination.
P =
  cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
  cond2#(false, x, y) -> ack#(x, y - 1) 
  cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
  cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
  ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
  ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
  f#(true, I12) -> f_1#(I12) 
  f#(true, I12) -> ack#(10, 10) 
  f_1#(I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
  ack#(I10, I11) -> cond1#(I10 = 0, I10, I11) [I10 >= 0 && I11 >= 0]
  ack#(I10, I11) -> cond2#(I11 = 0, I10, I11) [I10 >= 0 && I11 >= 0, not(I10 = 0)]
  ack#(I10, I11) -> ack#(I10 - 1, I11 + 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), I11 = 0]
  ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
  ack#(I10, I11) -> ack#(I10 - 1, ack(I10, I11 - 1)) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
  ackNat#(true, I8, I9) -> cond2#(I9 = 0, I8, I9) [not(I8 = 0)]
  ackNat#(true, I8, I9) -> ack#(I8 - 1, I9 + 1) [not(I8 = 0), I9 = 0]
  ackNat#(true, I8, I9) -> ack#(I8, I9 - 1) [not(I8 = 0), not(I9 = 0)]
  ackNat#(true, I8, I9) -> ack#(I8 - 1, ack(I8, I9 - 1)) [not(I8 = 0), not(I9 = 0)]
  cond1#(false, I2, I3) -> ack#(I2 - 1, ack(I2, I3 - 1)) [not(I3 = 0)]
  cond1#(false, I2, I3) -> ack#(I2, I3 - 1) [not(I3 = 0)]
  cond1#(false, I2, I3) -> ack#(I2 - 1, I3 + 1) [I3 = 0]
R = 
  cond2(false, x, y) -> ack(x - 1, ack(x, y - 1)) 
  cond2(true, I0, I1) -> ack(I0 - 1, I1 + 1) 
  cond1(false, I2, I3) -> cond2(I3 = 0, I2, I3) 
  cond1(true, I4, I5) -> I5 + 1 
  ackNat(false, I6, I7) -> 0 
  ackNat(true, I8, I9) -> cond1(I8 = 0, I8, I9) 
  ack(I10, I11) -> ackNat(I10 >= 0 && I11 >= 0, I10, I11) 
  f(true, I12) -> f(ack(10, 10) >= I12, I12 + 1) 

The dependency graph for this problem is:
  0 -> 13, 12, 11, 10, 9, 5
  1 -> 13, 12, 11, 10, 9, 5
  2 -> 13, 12, 11, 10, 9, 5
  3 -> 
  4 -> 
  5 -> 
  6 -> 8
  7 -> 13, 12, 10, 9, 5
  8 -> 7, 6
  9 -> 
  10 -> 
  11 -> 13, 12, 5, 9, 10
  12 -> 13, 12, 5, 9, 10, 11
  13 -> 13, 5, 9, 10, 11, 12
  14 -> 
  15 -> 5, 9, 10, 12, 13
  16 -> 5, 9, 10, 11, 12, 13
  17 -> 5, 9, 10, 11, 12, 13
  18 -> 5, 9, 10, 11, 12, 13
  19 -> 5, 9, 10, 11, 12, 13
  20 -> 5, 9, 10, 12, 13
Where:
  0) cond2#(false, x, y) -> ack#(x - 1, ack(x, y - 1)) 
  1) cond2#(false, x, y) -> ack#(x, y - 1) 
  2) cond2#(true, I0, I1) -> ack#(I0 - 1, I1 + 1) 
  3) cond1#(false, I2, I3) -> cond2#(I3 = 0, I2, I3) 
  4) ackNat#(true, I8, I9) -> cond1#(I8 = 0, I8, I9) 
  5) ack#(I10, I11) -> ackNat#(I10 >= 0 && I11 >= 0, I10, I11) 
  6) f#(true, I12) -> f_1#(I12) 
  7) f#(true, I12) -> ack#(10, 10) 
  8) f_1#(I12) -> f#(ack(10, 10) >= I12, I12 + 1) 
  9) ack#(I10, I11) -> cond1#(I10 = 0, I10, I11) [I10 >= 0 && I11 >= 0]
  10) ack#(I10, I11) -> cond2#(I11 = 0, I10, I11) [I10 >= 0 && I11 >= 0, not(I10 = 0)]
  11) ack#(I10, I11) -> ack#(I10 - 1, I11 + 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), I11 = 0]
  12) ack#(I10, I11) -> ack#(I10, I11 - 1) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
  13) ack#(I10, I11) -> ack#(I10 - 1, ack(I10, I11 - 1)) [I10 >= 0 && I11 >= 0, not(I10 = 0), not(I11 = 0)]
  14) ackNat#(true, I8, I9) -> cond2#(I9 = 0, I8, I9) [not(I8 = 0)]
  15) ackNat#(true, I8, I9) -> ack#(I8 - 1, I9 + 1) [not(I8 = 0), I9 = 0]
  16) ackNat#(true, I8, I9) -> ack#(I8, I9 - 1) [not(I8 = 0), not(I9 = 0)]
  17) ackNat#(true, I8, I9) -> ack#(I8 - 1, ack(I8, I9 - 1)) [not(I8 = 0), not(I9 = 0)]
  18) cond1#(false, I2, I3) -> ack#(I2 - 1, ack(I2, I3 - 1)) [not(I3 = 0)]
  19) cond1#(false, I2, I3) -> ack#(I2, I3 - 1) [not(I3 = 0)]

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job log

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actions all output return to Integer TRS Innermost