TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60000 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (55ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).
 | – Problem 5 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 6 was processed with processor BackwardInstantiation (2ms).
 |    |    | – Problem 7 was processed with processor Propagation (4ms).
 |    |    |    | – Problem 8 remains open; application of the following processors failed [ForwardNarrowing (0ms), BackwardInstantiation (0ms), ForwardInstantiation (0ms), Propagation (0ms)].

The following open problems remain:



Open Dependency Pair Problem 5

Dependency Pairs

logIter#(x, y)if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))if#(true, true, x, y)logIter#(x, y)

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

logarithm#(x)logIter#(x, 0)le#(s(x), s(y))le#(x, y)
logIter#(x, y)le#(s(0), x)logIter#(x, y)half#(x)
logIter#(x, y)inc#(y)half#(s(s(x)))half#(x)
logIter#(x, y)if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))if#(true, true, x, y)logIter#(x, y)
logIter#(x, y)le#(s(s(0)), x)inc#(s(x))inc#(x)

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

if#(true, true, x, y) → logIter#(x, y)logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))

half#(s(s(x))) → half#(x)

inc#(s(x)) → inc#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

half#(s(s(x)))half#(x)

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

half#(s(s(x)))half#(x)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

inc#(s(x))inc#(x)

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

inc#(s(x))inc#(x)

Problem 5: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(true, true, x, y)logIter#(x, y)logIter#(x, y)if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


Instantiation

For all potential predecessors l → r of the rule logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) on dependency pair chains it holds that: Thus, logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) is replaced by instances determined through the above matching. These instances are:
logIter#(_x, _y) → if#(le(s(0), _x), le(s(s(0)), _x), half(_x), inc(_y))

Problem 6: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(true, true, x, y)logIter#(x, y)logIter#(_x, _y)if#(le(s(0), _x), le(s(s(0)), _x), half(_x), inc(_y))

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


Instantiation

For all potential predecessors l → r of the rule logIter#(_x, _y) → if#(le(s(0), _x), le(s(s(0)), _x), half(_x), inc(_y)) on dependency pair chains it holds that: Thus, logIter#(_x, _y) → if#(le(s(0), _x), le(s(s(0)), _x), half(_x), inc(_y)) is replaced by instances determined through the above matching. These instances are:
logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))

Problem 7: Propagation



Dependency Pair Problem

Dependency Pairs

if#(true, true, x, y)logIter#(x, y)logIter#(x, y)if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))

Rewrite Rules

half(0)0half(s(0))0
half(s(s(x)))s(half(x))le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
inc(s(x))s(inc(x))inc(0)s(0)
logarithm(x)logIter(x, 0)logIter(x, y)if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y)logZeroErrorif(true, false, x, s(y))y
if(true, true, x, y)logIter(x, y)fg
fh

Original Signature

Termination of terms over the following signature is verified: f, g, half, true, logZeroError, h, logIter, 0, s, le, inc, logarithm, if, false

Strategy


The dependency pairs if#(true, true, x, y) → logIter#(x, y) and logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) are consolidated into the rule if#(true, true, x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) .

This is possible as

The dependency pairs if#(true, true, x, y) → logIter#(x, y) and logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) are consolidated into the rule if#(true, true, x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) .

This is possible as


Summary

Removed Dependency PairsAdded Dependency Pairs
logIter#(x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))if#(true, true, x, y) → if#(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if#(true, true, x, y) → logIter#(x, y)