YES

DP problem for innermost termination.
P =
  f13#(x1, x2, x3, x4, x5, x6, x7, x8) -> f12#(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4#(I24, I25, I26, I27, I28, I29, I30, I31) -> f8#(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4#(I32, I33, I34, I35, I36, I37, I38, I39) -> f7#(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7#(I40, I41, I42, I43, I44, I45, I46, I47) -> f5#(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9#(I56, I57, I58, I59, I60, I61, I62, I63) -> f3#(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5#(I96, I97, I98, I99, I100, I101, I102, I103) -> f7#(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f3#(I112, I113, I114, I115, I116, I117, I118, I119) -> f4#(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1#(I120, I121, I122, I123, I124, I125, I126, I127) -> f2#(I120, I121, I122, I123, I124, I125, I126, I127) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

The dependency graph for this problem is:
  0 -> 1
  1 -> 15
  2 -> 15
  3 -> 14
  4 -> 12
  5 -> 6
  6 -> 13
  7 -> 9, 10
  8 -> 14
  9 -> 11
  10 -> 11
  11 -> 12
  12 -> 7, 8
  13 -> 6
  14 -> 4, 5
  15 -> 2, 3
Where:
  0) f13#(x1, x2, x3, x4, x5, x6, x7, x8) -> f12#(x1, x2, x3, x4, x5, x6, x7, x8) 
  1) f12#(I0, I1, I2, I3, I4, I5, I6, I7) -> f1#(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  2) f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  3) f2#(I16, I17, I18, I19, I20, I21, I22, I23) -> f3#(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  4) f4#(I24, I25, I26, I27, I28, I29, I30, I31) -> f8#(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  5) f4#(I32, I33, I34, I35, I36, I37, I38, I39) -> f7#(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  6) f7#(I40, I41, I42, I43, I44, I45, I46, I47) -> f5#(I40, I41, I42, I43, I44, I45, I46, I47) 
  7) f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  8) f9#(I56, I57, I58, I59, I60, I61, I62, I63) -> f3#(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  9) f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  10) f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  11) f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  12) f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
  13) f5#(I96, I97, I98, I99, I100, I101, I102, I103) -> f7#(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  14) f3#(I112, I113, I114, I115, I116, I117, I118, I119) -> f4#(I112, I113, I114, I115, I116, I117, I118, I119) 
  15) f1#(I120, I121, I122, I123, I124, I125, I126, I127) -> f2#(I120, I121, I122, I123, I124, I125, I126, I127) 

We have the following SCCs.
  { 2, 15 }
  { 4, 7, 8, 9, 10, 11, 12, 14 }
  { 6, 13 }

DP problem for innermost termination.
P =
  f7#(I40, I41, I42, I43, I44, I45, I46, I47) -> f5#(I40, I41, I42, I43, I44, I45, I46, I47) 
  f5#(I96, I97, I98, I99, I100, I101, I102, I103) -> f7#(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

We use the reverse value criterion with the projection function NU:
  NU[f5#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 + z1 + -1 * (1 + z5)
  NU[f7#(z1,z2,z3,z4,z5,z6,z7,z8)] = -1 + z1 + -1 * (1 + z5)

This gives the following inequalities:
    ==>  -1 + I40 + -1 * (1 + I44) >= -1 + I40 + -1 * (1 + I44)
  1 + I100 <= -1 + I96  ==>  -1 + I96 + -1 * (1 + I100) > -1 + I96 + -1 * (1 + (1 + I100))  with  -1 + I96 + -1 * (1 + I100) >= 0

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f7#(I40, I41, I42, I43, I44, I45, I46, I47) -> f5#(I40, I41, I42, I43, I44, I45, I46, I47) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

The dependency graph for this problem is:
  6 -> 
Where:
  6) f7#(I40, I41, I42, I43, I44, I45, I46, I47) -> f5#(I40, I41, I42, I43, I44, I45, I46, I47) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f4#(I24, I25, I26, I27, I28, I29, I30, I31) -> f8#(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9#(I56, I57, I58, I59, I60, I61, I62, I63) -> f3#(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
  f3#(I112, I113, I114, I115, I116, I117, I118, I119) -> f4#(I112, I113, I114, I115, I116, I117, I118, I119) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

We use the extended value criterion with the projection function NU:
  NU[f10#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 3
  NU[f3#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 2
  NU[f11#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 3
  NU[f9#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 3
  NU[f8#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 3
  NU[f4#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x4 - 2

This gives the following inequalities:
  1 + I28 <= -1 + I24  ==>  I24 - I28 - 2 > I24 - I28 - 3  with  I24 - I28 - 2 >= 0
  1 + I50 <= I48  ==>  I48 - I52 - 3 >= I48 - I52 - 3
  rnd7 = rnd7 /\ I56 <= I58  ==>  I56 - I60 - 3 >= I56 - (1 + I60) - 2
    ==>  I64 - I68 - 3 >= I64 - I68 - 3
    ==>  I72 - I76 - 3 >= I72 - I76 - 3
    ==>  I80 - I84 - 3 >= I80 - I84 - 3
    ==>  I88 - I92 - 3 >= I88 - I92 - 3
    ==>  I112 - I116 - 2 >= I112 - I116 - 2

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9#(I56, I57, I58, I59, I60, I61, I62, I63) -> f3#(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
  f3#(I112, I113, I114, I115, I116, I117, I118, I119) -> f4#(I112, I113, I114, I115, I116, I117, I118, I119) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

The dependency graph for this problem is:
  7 -> 9, 10
  8 -> 14
  9 -> 11
  10 -> 11
  11 -> 12
  12 -> 7, 8
  14 -> 
Where:
  7) f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  8) f9#(I56, I57, I58, I59, I60, I61, I62, I63) -> f3#(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  9) f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  10) f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  11) f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  12) f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
  14) f3#(I112, I113, I114, I115, I116, I117, I118, I119) -> f4#(I112, I113, I114, I115, I116, I117, I118, I119) 

We have the following SCCs.
  { 7, 9, 10, 11, 12 }

DP problem for innermost termination.
P =
  f9#(I48, I49, I50, I51, I52, I53, I54, I55) -> f11#(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

We use the extended value criterion with the projection function NU:
  NU[f8#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x2 - 1
  NU[f10#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x2 - 2
  NU[f11#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x2 - 2
  NU[f9#(x0,x1,x2,x3,x4,x5,x6,x7)] = x0 - x2 - 1

This gives the following inequalities:
  1 + I50 <= I48  ==>  I48 - I50 - 1 > I48 - I50 - 2  with  I48 - I50 - 1 >= 0
    ==>  I64 - I66 - 2 >= I64 - I66 - 2
    ==>  I72 - I74 - 2 >= I72 - I74 - 2
    ==>  I80 - I82 - 2 >= I80 - (1 + I82) - 1
    ==>  I88 - I90 - 1 >= I88 - I90 - 1

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

The dependency graph for this problem is:
  9 -> 11
  10 -> 11
  11 -> 12
  12 -> 
Where:
  9) f11#(I64, I65, I66, I67, I68, I69, I70, I71) -> f10#(I64, I65, I66, I67, I68, I69, I70, I71) 
  10) f11#(I72, I73, I74, I75, I76, I77, I78, I79) -> f10#(I72, I73, I74, I75, I76, I74, I78, I79) 
  11) f10#(I80, I81, I82, I83, I84, I85, I86, I87) -> f8#(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  12) f8#(I88, I89, I90, I91, I92, I93, I94, I95) -> f9#(I88, I89, I90, I91, I92, I93, I94, I95) 

We have the following SCCs.
  

DP problem for innermost termination.
P =
  f2#(I8, I9, I10, I11, I12, I13, I14, I15) -> f1#(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f1#(I120, I121, I122, I123, I124, I125, I126, I127) -> f2#(I120, I121, I122, I123, I124, I125, I126, I127) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

We use the reverse value criterion with the projection function NU:
  NU[f1#(z1,z2,z3,z4,z5,z6,z7,z8)] = z2 + -1 * (1 + z4)
  NU[f2#(z1,z2,z3,z4,z5,z6,z7,z8)] = z2 + -1 * (1 + z4)

This gives the following inequalities:
  1 + I11 <= I9  ==>  I9 + -1 * (1 + I11) > I9 + -1 * (1 + (1 + I11))  with  I9 + -1 * (1 + I11) >= 0
    ==>  I121 + -1 * (1 + I123) >= I121 + -1 * (1 + I123)

We remove all the strictly oriented dependency pairs.

DP problem for innermost termination.
P =
  f1#(I120, I121, I122, I123, I124, I125, I126, I127) -> f2#(I120, I121, I122, I123, I124, I125, I126, I127) 
R = 
  f13(x1, x2, x3, x4, x5, x6, x7, x8) -> f12(x1, x2, x3, x4, x5, x6, x7, x8) 
  f12(I0, I1, I2, I3, I4, I5, I6, I7) -> f1(I0, I1, I2, 0, I4, I5, I6, rnd8) [rnd8 = rnd8]
  f2(I8, I9, I10, I11, I12, I13, I14, I15) -> f1(I8, I9, I10, 1 + I11, I12, I13, I14, I15) [1 + I11 <= I9]
  f2(I16, I17, I18, I19, I20, I21, I22, I23) -> f3(I17, I17, I18, I19, 0, I21, I22, I23) [I17 <= I19]
  f4(I24, I25, I26, I27, I28, I29, I30, I31) -> f8(I24, I25, 1 + I28, I27, I28, I28, I30, I31) [1 + I28 <= -1 + I24]
  f4(I32, I33, I34, I35, I36, I37, I38, I39) -> f7(I32, I33, I34, I35, 0, I37, I38, I39) [-1 + I32 <= I36]
  f7(I40, I41, I42, I43, I44, I45, I46, I47) -> f5(I40, I41, I42, I43, I44, I45, I46, I47) 
  f9(I48, I49, I50, I51, I52, I53, I54, I55) -> f11(I48, I49, I50, I51, I52, I53, I54, I55) [1 + I50 <= I48]
  f9(I56, I57, I58, I59, I60, I61, I62, I63) -> f3(I56, I57, I58, I59, 1 + I60, I61, rnd7, I63) [rnd7 = rnd7 /\ I56 <= I58]
  f11(I64, I65, I66, I67, I68, I69, I70, I71) -> f10(I64, I65, I66, I67, I68, I69, I70, I71) 
  f11(I72, I73, I74, I75, I76, I77, I78, I79) -> f10(I72, I73, I74, I75, I76, I74, I78, I79) 
  f10(I80, I81, I82, I83, I84, I85, I86, I87) -> f8(I80, I81, 1 + I82, I83, I84, I85, I86, I87) 
  f8(I88, I89, I90, I91, I92, I93, I94, I95) -> f9(I88, I89, I90, I91, I92, I93, I94, I95) 
  f5(I96, I97, I98, I99, I100, I101, I102, I103) -> f7(I96, I97, I98, I99, 1 + I100, I101, I102, I103) [1 + I100 <= -1 + I96]
  f5(I104, I105, I106, I107, I108, I109, I110, I111) -> f6(I104, I105, I106, I107, I108, I109, I110, I111) [-1 + I104 <= I108]
  f3(I112, I113, I114, I115, I116, I117, I118, I119) -> f4(I112, I113, I114, I115, I116, I117, I118, I119) 
  f1(I120, I121, I122, I123, I124, I125, I126, I127) -> f2(I120, I121, I122, I123, I124, I125, I126, I127) 

The dependency graph for this problem is:
  15 -> 
Where:
  15) f1#(I120, I121, I122, I123, I124, I125, I126, I127) -> f2#(I120, I121, I122, I123, I124, I125, I126, I127) 

We have the following SCCs.