TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60079 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (48ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor BackwardInstantiation (4ms).
 |    | – Problem 6 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (0ms), ForwardNarrowing (3ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (1ms)].
 | – Problem 5 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

if#(false, x, s(y), z)div#(minus(x, s(y)), s(y), z)div#(x, y, z)if#(lt(x, y), x, y, inc(z))

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: minus, division, 0, s, inc, if, div, false, true, lt


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(x, y, z)lt#(x, y)division#(x, y)div#(x, y, 0)
if#(false, x, s(y), z)div#(minus(x, s(y)), s(y), z)if#(false, x, s(y), z)minus#(x, s(y))
minus#(s(x), s(y))minus#(x, y)div#(x, y, z)inc#(z)
lt#(s(x), s(y))lt#(x, y)inc#(s(x))inc#(x)
div#(x, y, z)if#(lt(x, y), x, y, inc(z))

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt

Strategy


The following SCCs where found

if#(false, x, s(y), z) → div#(minus(x, s(y)), s(y), z)div#(x, y, z) → if#(lt(x, y), x, y, inc(z))

minus#(s(x), s(y)) → minus#(x, y)

inc#(s(x)) → inc#(x)

lt#(s(x), s(y)) → lt#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, 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)

Problem 4: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(false, x, s(y), z)div#(minus(x, s(y)), s(y), z)div#(x, y, z)if#(lt(x, y), x, y, inc(z))

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt

Strategy


Instantiation

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(s(x), s(y))minus#(x, y)

Rewrite Rules

division(x, y)div(x, y, 0)div(x, y, z)if(lt(x, y), x, y, inc(z))
if(true, x, y, z)zif(false, x, s(y), z)div(minus(x, s(y)), s(y), z)
minus(x, 0)xminus(s(x), s(y))minus(x, y)
lt(x, 0)falselt(0, s(y))true
lt(s(x), s(y))lt(x, y)inc(0)s(0)
inc(s(x))s(inc(x))

Original Signature

Termination of terms over the following signature is verified: 0, division, minus, inc, s, if, div, true, false, lt

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(s(x), s(y))minus#(x, y)