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 (32ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (148ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor ForwardNarrowing (3ms).
 |    | – Problem 5 was processed with processor ForwardNarrowing (2ms).
 |    |    | – Problem 6 was processed with processor ForwardNarrowing (1ms).
 |    |    |    | – Problem 7 was processed with processor ForwardInstantiation (3ms).
 |    |    |    |    | – Problem 8 was processed with processor ForwardInstantiation (1ms).
 |    |    |    |    |    | – Problem 9 was processed with processor ForwardInstantiation (1ms).
 |    |    |    |    |    |    | – Problem 10 remains open; application of the following processors failed [Propagation (0ms)].

The following open problems remain:



Open Dependency Pair Problem 4

Dependency Pairs

f#(x, s(y), b)f#(x, minus(s(y), s(0)), b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, div


Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

div#(s(x), s(y))minus#(x, y)f#(x, s(y), b)f#(x, minus(s(y), s(0)), b)
f#(x, s(y), b)div#(f(x, minus(s(y), s(0)), b), b)f#(x, s(y), b)minus#(s(y), s(0))
minus#(s(x), s(y))minus#(x, y)div#(s(x), s(y))div#(minus(x, y), s(y))

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


The following SCCs where found

f#(x, s(y), b) → f#(x, minus(s(y), s(0)), b)

minus#(s(x), s(y)) → minus#(x, y)

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

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


Polynomial Interpretation

Improved Usable rules

minus(s(x), s(y))minus(x, y)minus(x, 0)x
minus(0, x)0minus(x, x)0

The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 4: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

f#(x, s(y), b)f#(x, minus(s(y), s(0)), b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


The right-hand side of the rule f#(x, s(y), b) → f#(x, minus(s(y), s(0)), b) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(x, 0, b) 
f#(x, minus(_x32, 0), b) 
Thus, the rule f#(x, s(y), b) → f#(x, minus(s(y), s(0)), b) is replaced by the following rules:
f#(x, s(_x32), b) → f#(x, minus(_x32, 0), b)f#(x, s(0), b) → f#(x, 0, b)

Problem 5: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

f#(x, s(_x32), b)f#(x, minus(_x32, 0), b)f#(x, s(0), b)f#(x, 0, b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, div

Strategy


The right-hand side of the rule f#(x, s(_x32), b) → f#(x, minus(_x32, 0), b) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
f#(x, 0, b) 
f#(x, _x41, b) 
Thus, the rule f#(x, s(_x32), b) → f#(x, minus(_x32, 0), b) is replaced by the following rules:
f#(x, s(0), b) → f#(x, 0, b)f#(x, s(_x41), b) → f#(x, _x41, b)

Problem 6: ForwardNarrowing



Dependency Pair Problem

Dependency Pairs

f#(x, s(0), b)f#(x, 0, b)f#(x, s(_x41), b)f#(x, _x41, b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


The right-hand side of the rule f#(x, s(0), b) → f#(x, 0, b) is narrowed to the following relevant and irrelevant terms (a narrowing is irrelevant if by dropping it the correctness (and completeness) of the processor is not influenced).
Relevant TermsIrrelevant Terms
Thus, the rule f#(x, s(0), b) → f#(x, 0, b) is deleted.

Problem 7: ForwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(x, s(_x41), b)f#(x, _x41, b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, div

Strategy


Instantiation

For all potential successors l → r of the rule f#(x, s(_x41), b) → f#(x, _x41, b) on dependency pair chains it holds that: Thus, f#(x, s(_x41), b) → f#(x, _x41, b) is replaced by instances determined through the above matching. These instances are:
f#(x, s(s(__x41)), b) → f#(x, s(__x41), b)

Problem 8: ForwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(x, s(s(__x41)), b)f#(x, s(__x41), b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, s, div

Strategy


Instantiation

For all potential successors l → r of the rule f#(x, s(s(__x41)), b) → f#(x, s(__x41), b) on dependency pair chains it holds that: Thus, f#(x, s(s(__x41)), b) → f#(x, s(__x41), b) is replaced by instances determined through the above matching. These instances are:
f#(x, s(s(s(___x41))), b) → f#(x, s(s(___x41)), b)

Problem 9: ForwardInstantiation



Dependency Pair Problem

Dependency Pairs

f#(x, s(s(s(___x41))), b)f#(x, s(s(___x41)), b)

Rewrite Rules

minus(x, x)0minus(s(x), s(y))minus(x, y)
minus(0, x)0minus(x, 0)x
div(s(x), s(y))s(div(minus(x, y), s(y)))div(0, s(y))0
f(x, 0, b)xf(x, s(y), b)div(f(x, minus(s(y), s(0)), b), b)

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, div

Strategy


Instantiation

For all potential successors l → r of the rule f#(x, s(s(s(___x41))), b) → f#(x, s(s(___x41)), b) on dependency pair chains it holds that: Thus, f#(x, s(s(s(___x41))), b) → f#(x, s(s(___x41)), b) is replaced by instances determined through the above matching. These instances are:
f#(x, s(s(s(s(____x41)))), b) → f#(x, s(s(s(____x41))), b)