MAYBE

The TRS could not be proven terminating. The proof attempt took 1041 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (0ms).
 | – Problem 2 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (3ms), PolynomialLinearRange4iUR (170ms), DependencyGraph (2ms), PolynomialLinearRange8NegiUR (426ms), DependencyGraph (4ms), ReductionPairSAT (166ms), DependencyGraph (2ms), SizeChangePrinciple (17ms)].
 | – Problem 3 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 2

Dependency Pairs

if#(true, x, c)help#(x, s(c))help#(x, c)if#(lt(c, x), x, c)

Rewrite Rules

lt(0, s(x))truelt(x, 0)false
lt(s(x), s(y))lt(x, y)fac(x)help(x, 0)
help(x, c)if(lt(c, x), x, c)if(true, x, c)times(s(c), help(x, s(c)))
if(false, x, c)s(0)

Original Signature

Termination of terms over the following signature is verified: help, 0, s, times, if, fac, false, true, lt


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

if#(true, x, c)help#(x, s(c))lt#(s(x), s(y))lt#(x, y)
help#(x, c)lt#(c, x)fac#(x)help#(x, 0)
help#(x, c)if#(lt(c, x), x, c)

Rewrite Rules

lt(0, s(x))truelt(x, 0)false
lt(s(x), s(y))lt(x, y)fac(x)help(x, 0)
help(x, c)if(lt(c, x), x, c)if(true, x, c)times(s(c), help(x, s(c)))
if(false, x, c)s(0)

Original Signature

Termination of terms over the following signature is verified: help, 0, s, times, if, fac, true, false, lt

Strategy


The following SCCs where found

if#(true, x, c) → help#(x, s(c))help#(x, c) → if#(lt(c, x), x, c)

lt#(s(x), s(y)) → lt#(x, y)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

lt#(s(x), s(y))lt#(x, y)

Rewrite Rules

lt(0, s(x))truelt(x, 0)false
lt(s(x), s(y))lt(x, y)fac(x)help(x, 0)
help(x, c)if(lt(c, x), x, c)if(true, x, c)times(s(c), help(x, s(c)))
if(false, x, c)s(0)

Original Signature

Termination of terms over the following signature is verified: help, 0, s, times, if, fac, true, false, lt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

lt#(s(x), s(y))lt#(x, y)