TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60029 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (43ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 remains open; application of the following processors failed [SubtermCriterion (2ms), DependencyGraph (10ms), PolynomialLinearRange4iUR (496ms), DependencyGraph (7ms), PolynomialLinearRange8NegiUR (8188ms), DependencyGraph (46ms), ReductionPairSAT (timeout)].

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

if2#(true, x, y, z, u, v)quotIter#(x, y, z, 0, s(v))if2#(false, x, y, z, u, v)quotIter#(x, y, z, u, v)
if#(false, x, y, z, u, v)if2#(le(y, s(u)), x, y, s(z), s(u), v)quotIter#(x, s(y), z, u, v)if#(le(x, z), x, s(y), z, u, v)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)quot(x, 0)quotZeroErro
quot(x, s(y))quotIter(x, s(y), 0, 0, 0)quotIter(x, s(y), z, u, v)if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v)vif(false, x, y, z, u, v)if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v)quotIter(x, y, z, u, v)if2(true, x, y, z, u, v)quotIter(x, y, z, 0, s(v))

Original Signature

Termination of terms over the following signature is verified: 0, s, le, quotZeroErro, if, quotIter, false, true, if2, quot


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y))le#(x, y)if2#(true, x, y, z, u, v)quotIter#(x, y, z, 0, s(v))
quot#(x, s(y))quotIter#(x, s(y), 0, 0, 0)if2#(false, x, y, z, u, v)quotIter#(x, y, z, u, v)
if#(false, x, y, z, u, v)if2#(le(y, s(u)), x, y, s(z), s(u), v)if#(false, x, y, z, u, v)le#(y, s(u))
quotIter#(x, s(y), z, u, v)le#(x, z)quotIter#(x, s(y), z, u, v)if#(le(x, z), x, s(y), z, u, v)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)quot(x, 0)quotZeroErro
quot(x, s(y))quotIter(x, s(y), 0, 0, 0)quotIter(x, s(y), z, u, v)if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v)vif(false, x, y, z, u, v)if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v)quotIter(x, y, z, u, v)if2(true, x, y, z, u, v)quotIter(x, y, z, 0, s(v))

Original Signature

Termination of terms over the following signature is verified: 0, le, s, quotZeroErro, quotIter, if, true, false, if2, quot

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

if2#(true, x, y, z, u, v) → quotIter#(x, y, z, 0, s(v))if2#(false, x, y, z, u, v) → quotIter#(x, y, z, u, v)
if#(false, x, y, z, u, v) → if2#(le(y, s(u)), x, y, s(z), s(u), v)quotIter#(x, s(y), z, u, v) → if#(le(x, z), x, s(y), z, u, v)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y))le#(x, y)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)quot(x, 0)quotZeroErro
quot(x, s(y))quotIter(x, s(y), 0, 0, 0)quotIter(x, s(y), z, u, v)if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v)vif(false, x, y, z, u, v)if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v)quotIter(x, y, z, u, v)if2(true, x, y, z, u, v)quotIter(x, y, z, 0, s(v))

Original Signature

Termination of terms over the following signature is verified: 0, le, s, quotZeroErro, quotIter, if, true, false, if2, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y))le#(x, y)