TIMEOUT

The TRS could not be proven terminating. The proof attempt took 60028 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (45ms).
 | – Problem 2 was processed with processor SubtermCriterion (2ms).
 | – Problem 3 was processed with processor BackwardInstantiation (4ms).
 |    | – Problem 5 remains open; application of the following processors failed [ForwardInstantiation (2ms), Propagation (4ms), ForwardNarrowing (1ms), BackwardInstantiation (2ms), ForwardInstantiation (3ms), Propagation (3ms)].
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

The following open problems remain:



Open Dependency Pair Problem 3

Dependency Pairs

cond2#(false, x, y)diff#(s(x), y)cond1#(false, x, y)cond2#(gt(x, y), x, y)
cond2#(true, x, y)diff#(x, s(y))diff#(x, y)cond1#(equal(x, y), x, y)

Rewrite Rules

diff(x, y)cond1(equal(x, y), x, y)cond1(true, x, y)0
cond1(false, x, y)cond2(gt(x, y), x, y)cond2(true, x, y)s(diff(x, s(y)))
cond2(false, x, y)s(diff(s(x), y))gt(0, v)false
gt(s(u), 0)truegt(s(u), s(v))gt(u, v)
equal(0, 0)trueequal(s(x), 0)false
equal(0, s(y))falseequal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, diff, false, true, gt, equal, cond1


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

cond2#(false, x, y)diff#(s(x), y)cond1#(false, x, y)cond2#(gt(x, y), x, y)
cond1#(false, x, y)gt#(x, y)equal#(s(x), s(y))equal#(x, y)
cond2#(true, x, y)diff#(x, s(y))gt#(s(u), s(v))gt#(u, v)
diff#(x, y)cond1#(equal(x, y), x, y)diff#(x, y)equal#(x, y)

Rewrite Rules

diff(x, y)cond1(equal(x, y), x, y)cond1(true, x, y)0
cond1(false, x, y)cond2(gt(x, y), x, y)cond2(true, x, y)s(diff(x, s(y)))
cond2(false, x, y)s(diff(s(x), y))gt(0, v)false
gt(s(u), 0)truegt(s(u), s(v))gt(u, v)
equal(0, 0)trueequal(s(x), 0)false
equal(0, s(y))falseequal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, diff, true, false, equal, gt, cond1

Strategy


The following SCCs where found

equal#(s(x), s(y)) → equal#(x, y)

cond2#(false, x, y) → diff#(s(x), y)cond1#(false, x, y) → cond2#(gt(x, y), x, y)
cond2#(true, x, y) → diff#(x, s(y))diff#(x, y) → cond1#(equal(x, y), x, y)

gt#(s(u), s(v)) → gt#(u, v)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

equal#(s(x), s(y))equal#(x, y)

Rewrite Rules

diff(x, y)cond1(equal(x, y), x, y)cond1(true, x, y)0
cond1(false, x, y)cond2(gt(x, y), x, y)cond2(true, x, y)s(diff(x, s(y)))
cond2(false, x, y)s(diff(s(x), y))gt(0, v)false
gt(s(u), 0)truegt(s(u), s(v))gt(u, v)
equal(0, 0)trueequal(s(x), 0)false
equal(0, s(y))falseequal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, diff, true, false, equal, gt, cond1

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

equal#(s(x), s(y))equal#(x, y)

Problem 3: BackwardInstantiation



Dependency Pair Problem

Dependency Pairs

cond2#(false, x, y)diff#(s(x), y)cond1#(false, x, y)cond2#(gt(x, y), x, y)
cond2#(true, x, y)diff#(x, s(y))diff#(x, y)cond1#(equal(x, y), x, y)

Rewrite Rules

diff(x, y)cond1(equal(x, y), x, y)cond1(true, x, y)0
cond1(false, x, y)cond2(gt(x, y), x, y)cond2(true, x, y)s(diff(x, s(y)))
cond2(false, x, y)s(diff(s(x), y))gt(0, v)false
gt(s(u), 0)truegt(s(u), s(v))gt(u, v)
equal(0, 0)trueequal(s(x), 0)false
equal(0, s(y))falseequal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, diff, true, false, equal, gt, cond1

Strategy


Instantiation

For all potential predecessors l → r of the rule diff#(x, y) → cond1#(equal(x, y), x, y) on dependency pair chains it holds that: Thus, diff#(x, y) → cond1#(equal(x, y), x, y) is replaced by instances determined through the above matching. These instances are:
diff#(s(_x), _y) → cond1#(equal(s(_x), _y), s(_x), _y)diff#(_x, s(_y)) → cond1#(equal(_x, s(_y)), _x, s(_y))

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

gt#(s(u), s(v))gt#(u, v)

Rewrite Rules

diff(x, y)cond1(equal(x, y), x, y)cond1(true, x, y)0
cond1(false, x, y)cond2(gt(x, y), x, y)cond2(true, x, y)s(diff(x, s(y)))
cond2(false, x, y)s(diff(s(x), y))gt(0, v)false
gt(s(u), 0)truegt(s(u), s(v))gt(u, v)
equal(0, 0)trueequal(s(x), 0)false
equal(0, s(y))falseequal(s(x), s(y))equal(x, y)

Original Signature

Termination of terms over the following signature is verified: cond2, 0, s, diff, true, false, equal, gt, cond1

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

gt#(s(u), s(v))gt#(u, v)