YES

The TRS could be proven terminating. The proof took 424 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (64ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).
 |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (211ms).
 | – Problem 4 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

f#(s(x), s(y), z, u)le#(x, y)le#(s(x), s(y))le#(x, y)
f#(s(x), s(y), z, u)f#(s(x), minus(y, x), z, u)minus#(s(x), s(y))minus#(x, y)
f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)f#(s(x), s(y), z, u)minus#(y, x)
f#(s(x), s(y), z, u)f#(x, u, z, u)f#(s(x), s(y), z, u)if#(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))
f#(s(x), 0, z, u)minus#(z, s(x))perfectp#(s(x))f#(x, s(0), s(x), s(x))

Rewrite Rules

minus(0, y)0minus(s(x), 0)s(x)
minus(s(x), s(y))minus(x, y)le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
if(true, x, y)xif(false, x, y)y
perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false

Strategy


The following SCCs where found

le#(s(x), s(y)) → le#(x, y)

minus#(s(x), s(y)) → minus#(x, y)

f#(s(x), s(y), z, u) → f#(s(x), minus(y, x), z, u)f#(s(x), s(y), z, u) → f#(x, u, z, u)
f#(s(x), 0, z, u) → f#(x, u, minus(z, s(x)), u)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

minus(0, y)0minus(s(x), 0)s(x)
minus(s(x), s(y))minus(x, y)le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
if(true, x, y)xif(false, x, y)y
perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(s(x), s(y), z, u)f#(s(x), minus(y, x), z, u)f#(s(x), s(y), z, u)f#(x, u, z, u)
f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)

Rewrite Rules

minus(0, y)0minus(s(x), 0)s(x)
minus(s(x), s(y))minus(x, y)le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
if(true, x, y)xif(false, x, y)y
perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(s(x), s(y), z, u)f#(x, u, z, u)f#(s(x), 0, z, u)f#(x, u, minus(z, s(x)), u)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(s(x), s(y), z, u)f#(s(x), minus(y, x), z, u)

Rewrite Rules

minus(0, y)0minus(s(x), 0)s(x)
minus(s(x), s(y))minus(x, y)le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
if(true, x, y)xif(false, x, y)y
perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, 0, minus, le, s, if, perfectp, false, true

Strategy


Polynomial Interpretation

Improved Usable rules

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

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

f#(s(x), s(y), z, u)f#(s(x), minus(y, x), z, u)

Problem 4: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

le#(s(x), s(y))le#(x, y)

Rewrite Rules

minus(0, y)0minus(s(x), 0)s(x)
minus(s(x), s(y))minus(x, y)le(0, y)true
le(s(x), 0)falsele(s(x), s(y))le(x, y)
if(true, x, y)xif(false, x, y)y
perfectp(0)falseperfectp(s(x))f(x, s(0), s(x), s(x))
f(0, y, 0, u)truef(0, y, s(z), u)false
f(s(x), 0, z, u)f(x, u, minus(z, s(x)), u)f(s(x), s(y), z, u)if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u))

Original Signature

Termination of terms over the following signature is verified: f, minus, 0, s, le, if, perfectp, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

le#(s(x), s(y))le#(x, y)