TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60001 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (67ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (0ms).
 | – Problem 4 was processed with processor BackwardInstantiation (2ms).
 |    | – Problem 6 remains open; application of the following processors failed [ForwardInstantiation (1ms), Propagation (1ms), ForwardNarrowing (1ms), BackwardInstantiation (1ms), ForwardInstantiation (1ms), Propagation (3ms)].
 | – Problem 5 was processed with processor SubtermCriterion (0ms).

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

if#(true, x, s(y), c)help#(x, s(y), plus(c, s(y)))help#(x, s(y), c)if#(le(c, x), x, s(y), c)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, minus, 0, s, le, if, mod, false, true


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y))le#(x, y)plus#(x, s(y))plus#(x, y)
mod#(x, s(y))help#(x, s(y), 0)minus#(s(x), s(y))minus#(x, y)
if#(true, x, s(y), c)help#(x, s(y), plus(c, s(y)))if#(false, x, s(y), c)minus#(x, minus(c, s(y)))
help#(x, s(y), c)if#(le(c, x), x, s(y), c)if#(false, x, s(y), c)minus#(c, s(y))
help#(x, s(y), c)le#(c, x)if#(true, x, s(y), c)plus#(c, s(y))

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

if#(true, x, s(y), c) → help#(x, s(y), plus(c, s(y)))help#(x, s(y), c) → if#(le(c, x), x, s(y), c)

plus#(x, s(y)) → plus#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

plus#(x, s(y))plus#(x, y)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

plus#(x, s(y))plus#(x, y)

Problem 3: 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)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, 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: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

if#(true, x, s(y), c)help#(x, s(y), plus(c, s(y)))help#(x, s(y), c)if#(le(c, x), x, s(y), c)

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false

Strategy


Instantiation

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

Problem 5: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

le(0, y)truele(s(x), 0)false
le(s(x), s(y))le(x, y)minus(x, 0)x
minus(0, s(y))0minus(s(x), s(y))minus(x, y)
plus(x, 0)xplus(x, s(y))s(plus(x, y))
mod(s(x), 0)0mod(x, s(y))help(x, s(y), 0)
help(x, s(y), c)if(le(c, x), x, s(y), c)if(true, x, s(y), c)help(x, s(y), plus(c, s(y)))
if(false, x, s(y), c)minus(x, minus(c, s(y)))

Original Signature

Termination of terms over the following signature is verified: plus, help, 0, minus, le, s, if, mod, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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