YES

The TRS could be proven terminating. The proof took 868 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (16ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (234ms).
 | – Problem 4 was processed with processor PolynomialLinearRange4iUR (565ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

quot#(s(x), s(y))minus#(x, y)log#(s(s(x)))quot#(x, s(s(0)))
minus#(x, s(y))minus#(x, y)quot#(s(x), s(y))quot#(minus(x, y), s(y))
minus#(x, s(y))pred#(minus(x, y))log#(s(s(x)))log#(s(quot(x, s(s(0)))))

Rewrite Rules

pred(s(x))xminus(x, 0)x
minus(x, s(y))pred(minus(x, y))quot(0, s(y))0
quot(s(x), s(y))s(quot(minus(x, y), s(y)))log(s(0))0
log(s(s(x)))s(log(s(quot(x, s(s(0))))))

Original Signature

Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log

Strategy


The following SCCs where found

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

minus#(x, s(y)) → minus#(x, y)

log#(s(s(x))) → log#(s(quot(x, s(s(0)))))

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

pred(s(x))xminus(x, 0)x
minus(x, s(y))pred(minus(x, y))quot(0, s(y))0
quot(s(x), s(y))s(quot(minus(x, y), s(y)))log(s(0))0
log(s(s(x)))s(log(s(quot(x, s(s(0))))))

Original Signature

Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

pred(s(x))xminus(x, 0)x
minus(x, s(y))pred(minus(x, y))quot(0, s(y))0
quot(s(x), s(y))s(quot(minus(x, y), s(y)))log(s(0))0
log(s(s(x)))s(log(s(quot(x, s(s(0))))))

Original Signature

Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

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

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

log#(s(s(x)))log#(s(quot(x, s(s(0)))))

Rewrite Rules

pred(s(x))xminus(x, 0)x
minus(x, s(y))pred(minus(x, y))quot(0, s(y))0
quot(s(x), s(y))s(quot(minus(x, y), s(y)))log(s(0))0
log(s(s(x)))s(log(s(quot(x, s(s(0))))))

Original Signature

Termination of terms over the following signature is verified: minus, 0, s, pred, quot, log

Strategy


Polynomial Interpretation

Improved Usable rules

pred(s(x))xquot(0, s(y))0
quot(s(x), s(y))s(quot(minus(x, y), s(y)))minus(x, 0)x
minus(x, s(y))pred(minus(x, y))

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

log#(s(s(x)))log#(s(quot(x, s(s(0)))))