TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60043 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (51ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (2ms), ForwardNarrowing (0ms), BackwardInstantiation (2ms), ForwardInstantiation (1ms), Propagation (1ms)].

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

div#(x, y, z)if#(ge(y, s(0)), ge(x, y), x, y, z)if#(true, true, x, y, z)div#(minus(x, y), y, id_inc(z))

Rewrite Rules

ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)minus(x, 0)x
minus(0, y)0minus(s(x), s(y))minus(x, y)
id_inc(x)xid_inc(x)s(x)
quot(x, y)div(x, y, 0)div(x, y, z)if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z)div_by_zeroif(true, false, x, y, z)z
if(true, true, x, y, z)div(minus(x, y), y, id_inc(z))

Original Signature

Termination of terms over the following signature is verified: div_by_zero, id_inc, minus, 0, s, if, div, false, true, ge, quot


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(x, y, z)ge#(x, y)div#(x, y, z)if#(ge(y, s(0)), ge(x, y), x, y, z)
div#(x, y, z)ge#(y, s(0))ge#(s(x), s(y))ge#(x, y)
minus#(s(x), s(y))minus#(x, y)if#(true, true, x, y, z)minus#(x, y)
if#(true, true, x, y, z)div#(minus(x, y), y, id_inc(z))quot#(x, y)div#(x, y, 0)
if#(true, true, x, y, z)id_inc#(z)

Rewrite Rules

ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)minus(x, 0)x
minus(0, y)0minus(s(x), s(y))minus(x, y)
id_inc(x)xid_inc(x)s(x)
quot(x, y)div(x, y, 0)div(x, y, z)if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z)div_by_zeroif(true, false, x, y, z)z
if(true, true, x, y, z)div(minus(x, y), y, id_inc(z))

Original Signature

Termination of terms over the following signature is verified: div_by_zero, 0, minus, id_inc, s, if, div, true, false, ge, quot

Strategy


The following SCCs where found

minus#(s(x), s(y)) → minus#(x, y)

ge#(s(x), s(y)) → ge#(x, y)

div#(x, y, z) → if#(ge(y, s(0)), ge(x, y), x, y, z)if#(true, true, x, y, z) → div#(minus(x, y), y, id_inc(z))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)minus(x, 0)x
minus(0, y)0minus(s(x), s(y))minus(x, y)
id_inc(x)xid_inc(x)s(x)
quot(x, y)div(x, y, 0)div(x, y, z)if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z)div_by_zeroif(true, false, x, y, z)z
if(true, true, x, y, z)div(minus(x, y), y, id_inc(z))

Original Signature

Termination of terms over the following signature is verified: div_by_zero, 0, minus, id_inc, s, if, div, true, false, ge, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

ge#(s(x), s(y))ge#(x, y)

Rewrite Rules

ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)minus(x, 0)x
minus(0, y)0minus(s(x), s(y))minus(x, y)
id_inc(x)xid_inc(x)s(x)
quot(x, y)div(x, y, 0)div(x, y, z)if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z)div_by_zeroif(true, false, x, y, z)z
if(true, true, x, y, z)div(minus(x, y), y, id_inc(z))

Original Signature

Termination of terms over the following signature is verified: div_by_zero, 0, minus, id_inc, s, if, div, true, false, ge, quot

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

ge#(s(x), s(y))ge#(x, y)

Problem 4: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

div#(x, y, z)if#(ge(y, s(0)), ge(x, y), x, y, z)if#(true, true, x, y, z)div#(minus(x, y), y, id_inc(z))

Rewrite Rules

ge(x, 0)truege(0, s(y))false
ge(s(x), s(y))ge(x, y)minus(x, 0)x
minus(0, y)0minus(s(x), s(y))minus(x, y)
id_inc(x)xid_inc(x)s(x)
quot(x, y)div(x, y, 0)div(x, y, z)if(ge(y, s(0)), ge(x, y), x, y, z)
if(false, b, x, y, z)div_by_zeroif(true, false, x, y, z)z
if(true, true, x, y, z)div(minus(x, y), y, id_inc(z))

Original Signature

Termination of terms over the following signature is verified: div_by_zero, 0, minus, id_inc, s, if, div, true, false, ge, quot

Strategy


Instantiation

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