Integ TRS Inner 43557 pair #381740682

details
property	value
status	complete
benchmark	a.03.itrs
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n029.star.cs.uiowa.edu
space	Mixed_ITRS_2014
run statistics
property	value
solver	Ctrl
configuration	Itrs
runtime (wallclock)	23.8304510117 seconds
cpu usage	25.315096026
max memory	2.721792E7
stage attributes
key	value
output-size	13103
starexec-result	YES
output
/export/starexec/sandbox/solver/bin/starexec_run_Itrs /export/starexec/sandbox/benchmark/theBenchmark.itrs /export/starexec/sandbox/output/output_files


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YES

DP problem for innermost termination.
P =
  eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2]
  eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2]
  eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1]
  eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
  eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1]
  eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1]
  eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21]
  eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25]
  eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2]
  eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34]
  eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2]
R = 
  eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2]
  eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2]
  eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1]
  eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
  eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1]
  eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1]
  eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21]
  eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25]
  eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2]
  eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34]
  eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2]

The dependency graph for this problem is:
  0 -> 8
  1 -> 8
  2 -> 
  3 -> 0, 1
  4 -> 2, 3, 4, 5, 6, 7
  5 -> 2, 3, 4, 5, 6, 7
  6 -> 0, 1
  7 -> 0, 1
  8 -> 2, 3, 4, 5, 6, 7
  9 -> 8
  10 -> 8
Where:
  0) eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2]
  1) eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2]
  2) eval_3#(I4, I5, I6, I7, B1) -> eval_3#(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1]
  3) eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
  4) eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1]
  5) eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1]
  6) eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21]
  7) eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25]
  8) eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2]
  9) eval_1#(I32, I33, I34, I35, B8) -> eval_2#(I32, I33, I34, I35 - 1, B8) [2 > I34]
  10) eval_1#(I36, I37, I38, I39, B9) -> eval_2#(I36, I37, I38 - 1, I39, B9) [I38 >= 2]

We have the following SCCs.
  { 0, 1, 3, 4, 5, 6, 7, 8 }

DP problem for innermost termination.
P =
  eval_4#(i, j, l, r, n) -> eval_2#(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2]
  eval_4#(I0, I1, I2, I3, B0) -> eval_2#(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2]
  eval_3#(I8, I9, I10, I11, B2) -> eval_4#(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
  eval_3#(I12, I13, I14, I15, B3) -> eval_3#(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1]
  eval_3#(I16, I17, I18, I19, B4) -> eval_3#(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1]
  eval_3#(I20, I21, I22, I23, B5) -> eval_4#(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21]
  eval_3#(I24, I25, I26, I27, B6) -> eval_4#(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25]
  eval_2#(I28, I29, I30, I31, B7) -> eval_3#(I30, 2 * I30, I30, I31, B7) [I31 >= 2]
R = 
  eval_4(i, j, l, r, n) -> eval_2(i, j, l, r - 1, n) [2 > l && l >= 1 && r >= 2]
  eval_4(I0, I1, I2, I3, B0) -> eval_2(I0, I1, I2 - 1, I3, B0) [I2 >= 2 && I2 >= 1 && I3 >= 2]
  eval_3(I4, I5, I6, I7, B1) -> eval_3(I5, 2 * I5, I6, I7, B1) [I7 >= I5 && I5 > I7 - 1 && I5 >= 1]
  eval_3(I8, I9, I10, I11, B2) -> eval_4(I8, I9, I10, I11, B2) [I11 >= I9 && I9 > I11 - 1]
  eval_3(I12, I13, I14, I15, B3) -> eval_3(I13 + 1, 2 * I13 + 2, I14, I15, B3) [I15 >= I13 && I15 - 1 >= I13 && I13 >= 1]
  eval_3(I16, I17, I18, I19, B4) -> eval_3(I17, 2 * I17, I18, I19, B4) [I19 >= I17 && I19 - 1 >= I17 && I17 >= 1]
  eval_3(I20, I21, I22, I23, B5) -> eval_4(I20, I21 + 1, I22, I23, B5) [I23 >= I21 && I23 - 1 >= I21]
  eval_3(I24, I25, I26, I27, B6) -> eval_4(I24, I25, I26, I27, B6) [I27 >= I25 && I27 - 1 >= I25]
  eval_2(I28, I29, I30, I31, B7) -> eval_3(I30, 2 * I30, I30, I31, B7) [I31 >= 2]
  eval_1(I32, I33, I34, I35, B8) -> eval_2(I32, I33, I34, I35 - 1, B8) [2 > I34]
  eval_1(I36, I37, I38, I39, B9) -> eval_2(I36, I37, I38 - 1, I39, B9) [I38 >= 2]

We use the reverse value criterion with the projection function NU:
  NU[eval_3#(z1,z2,z3,z4,z5)] = z3
  NU[eval_2#(z1,z2,z3,z4,z5)] = z3
  NU[eval_4#(z1,z2,z3,z4,z5)] = z3

This gives the following inequalities:
  2 > l && l >= 1 && r >= 2  ==>  l >= l
  I2 >= 2 && I2 >= 1 && I3 >= 2  ==>  I2 > I2 - 1  with  I2 >= 0
  I11 >= I9 && I9 > I11 - 1  ==>  I10 >= I10
  I15 >= I13 && I15 - 1 >= I13 && I13 >= 1  ==>  I14 >= I14
  I19 >= I17 && I19 - 1 >= I17 && I17 >= 1  ==>  I18 >= I18
  I23 >= I21 && I23 - 1 >= I21  ==>  I22 >= I22
  I27 >= I25 && I27 - 1 >= I25  ==>  I26 >= I26
  I31 >= 2  ==>  I30 >= I30
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