YES

The TRS could be proven terminating. The proof took 505 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (27ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (299ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

mod#(s(x), s(y)) → le#(y, x)le#(s(x), s(y)) → le#(x, y)
mod#(s(x), s(y)) → if_mod#(le(y, x), s(x), s(y))if_mod#(true, s(x), s(y)) → minus#(x, y)
minus#(x, s(y)) → minus#(x, y)minus#(x, s(y)) → pred#(minus(x, y))
if_mod#(true, s(x), s(y)) → mod#(minus(x, y), s(y))

Rewrite Rules

le(0, y) → truele(s(x), 0) → false
le(s(x), s(y)) → le(x, y)pred(s(x)) → x
minus(x, 0) → xminus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, pred, if_mod

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

minus#(x, s(y)) → minus#(x, y)

mod#(s(x), s(y)) → if_mod#(le(y, x), s(x), s(y))if_mod#(true, s(x), s(y)) → mod#(minus(x, y), s(y))

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

mod#(s(x), s(y)) → if_mod#(le(y, x), s(x), s(y))if_mod#(true, s(x), s(y)) → mod#(minus(x, y), s(y))

Rewrite Rules

le(0, y) → truele(s(x), 0) → false
le(s(x), s(y)) → le(x, y)pred(s(x)) → x
minus(x, 0) → xminus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, pred, if_mod

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:

mod#(s(x), s(y)) → if_mod#(le(y, x), s(x), s(y))if_mod#(true, s(x), s(y)) → mod#(minus(x, y), s(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)pred(s(x)) → x
minus(x, 0) → xminus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, pred, if_mod

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y)) → le#(x, y)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

minus#(x, s(y)) → minus#(x, y)

Rewrite Rules

le(0, y) → truele(s(x), 0) → false
le(s(x), s(y)) → le(x, y)pred(s(x)) → x
minus(x, 0) → xminus(x, s(y)) → pred(minus(x, y))
mod(0, y) → 0mod(s(x), 0) → 0
mod(s(x), s(y)) → if_mod(le(y, x), s(x), s(y))if_mod(true, s(x), s(y)) → mod(minus(x, y), s(y))
if_mod(false, s(x), s(y)) → s(x)

Original Signature

Termination of terms over the following signature is verified: 0, minus, le, s, mod, true, false, pred, if_mod

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

minus#(x, s(y)) → minus#(x, y)