YES

The TRS could be proven terminating. The proof took 367 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (68ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4 (62ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4 (110ms).
 |    | – Problem 4 was processed with processor DependencyGraph (0ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

U12#(y', y, x, x')plus#(x', y')plus#(x, y)U01#(x, y, x)
U11#(s(x'), y, x)T(y)plus#(x, y)U11#(x, y, x)
fib#(s(x))U31#(fib(x), x)fib#(s(x))fib#(x)
U11#(s(x'), y, x)U12#(y, y, x, x')U01#(0, y, x)U02#(y, y, x)
U12#(y', y, x, x')T(x')U01#(0, y, x)T(y)
U31#(pair(y, z), x)plus#(y, z)

Rewrite Rules

plus(x, y)U01(x, y, x)U01(0, y, x)U02(y, y, x)
U02(y', y, x)y'plus(x, y)U11(x, y, x)
U11(s(x'), y, x)U12(y, y, x, x')U12(y', y, x, x')s(plus(x', y'))
fib(0)pair(0, s(0))fib(s(x))U31(fib(x), x)
U31(pair(y, z), x)pair(z, plus(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, pair, fib

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(U11#) = μ(U12#) = μ(U31#) = μ(fib) = μ(s) = μ(U01) = μ(U02) = μ(U11) = μ(U12) = μ(fib#) = μ(U31) = μ(U02#) = μ(U01#) = {1}
μ(plus) = μ(pair) = μ(plus#) = {1, 2}


The following SCCs where found

U12#(y', y, x, x') → plus#(x', y')plus#(x, y) → U11#(x, y, x)
U11#(s(x'), y, x) → U12#(y, y, x, x')

fib#(s(x)) → fib#(x)

Problem 2: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

fib#(s(x))fib#(x)

Rewrite Rules

plus(x, y)U01(x, y, x)U01(0, y, x)U02(y, y, x)
U02(y', y, x)y'plus(x, y)U11(x, y, x)
U11(s(x'), y, x)U12(y, y, x, x')U12(y', y, x, x')s(plus(x', y'))
fib(0)pair(0, s(0))fib(s(x))U31(fib(x), x)
U31(pair(y, z), x)pair(z, plus(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, pair, fib

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(U11#) = μ(U12#) = μ(U31#) = μ(fib) = μ(s) = μ(U01) = μ(U02) = μ(U11) = μ(U12) = μ(U31) = μ(fib#) = μ(U02#) = μ(U01#) = {1}
μ(plus) = μ(pair) = μ(plus#) = {1, 2}


Polynomial Interpretation

There are no usable rules

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

fib#(s(x))fib#(x)

Problem 3: PolynomialLinearRange4



Dependency Pair Problem

Dependency Pairs

U12#(y', y, x, x')plus#(x', y')plus#(x, y)U11#(x, y, x)
U11#(s(x'), y, x)U12#(y, y, x, x')

Rewrite Rules

plus(x, y)U01(x, y, x)U01(0, y, x)U02(y, y, x)
U02(y', y, x)y'plus(x, y)U11(x, y, x)
U11(s(x'), y, x)U12(y, y, x, x')U12(y', y, x, x')s(plus(x', y'))
fib(0)pair(0, s(0))fib(s(x))U31(fib(x), x)
U31(pair(y, z), x)pair(z, plus(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, pair, fib

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(U11#) = μ(U12#) = μ(U31#) = μ(fib) = μ(s) = μ(U01) = μ(U02) = μ(U11) = μ(U12) = μ(U31) = μ(fib#) = μ(U02#) = μ(U01#) = {1}
μ(plus) = μ(pair) = μ(plus#) = {1, 2}


Polynomial Interpretation

Standard Usable rules

fib(s(x))U31(fib(x), x)U12(y', y, x, x')s(plus(x', y'))
plus(x, y)U01(x, y, x)U01(0, y, x)U02(y, y, x)
U02(y', y, x)y'U31(pair(y, z), x)pair(z, plus(y, z))
fib(0)pair(0, s(0))U11(s(x'), y, x)U12(y, y, x, x')
plus(x, y)U11(x, y, x)

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

U12#(y', y, x, x')plus#(x', y')U11#(s(x'), y, x)U12#(y, y, x, x')

Problem 4: DependencyGraph



Dependency Pair Problem

Dependency Pairs

plus#(x, y)U11#(x, y, x)

Rewrite Rules

plus(x, y)U01(x, y, x)U01(0, y, x)U02(y, y, x)
U02(y', y, x)y'plus(x, y)U11(x, y, x)
U11(s(x'), y, x)U12(y, y, x, x')U12(y', y, x, x')s(plus(x', y'))
fib(0)pair(0, s(0))fib(s(x))U31(fib(x), x)
U31(pair(y, z), x)pair(z, plus(y, z))

Original Signature

Termination of terms over the following signature is verified: plus, 0, s, pair, fib

Strategy

Context-sensitive strategy:
μ(T) = μ(0) = ∅
μ(U11#) = μ(U12#) = μ(U31#) = μ(fib) = μ(s) = μ(U01) = μ(U02) = μ(U11) = μ(U12) = μ(fib#) = μ(U31) = μ(U02#) = μ(U01#) = {1}
μ(plus) = μ(pair) = μ(plus#) = {1, 2}


There are no SCCs!