YES

The TRS could be proven terminating. The proof took 1663 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (19ms).
 | – Problem 2 was processed with processor SubtermCriterion (3ms).
 | – Problem 3 was processed with processor PolynomialLinearRange4iUR (691ms).
 |    | – Problem 4 was processed with processor PolynomialLinearRange4iUR (542ms).
 |    |    | – Problem 5 was processed with processor PolynomialLinearRange4iUR (339ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

g#(x, c(y)) → g#(s(x), y)g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))
g#(x, c(y)) → if#(f(x), c(g(s(x), y)), c(y))g#(x, c(y)) → g#(x, y)
g#(x, c(y)) → f#(x)f#(s(x)) → f#(x)

Rewrite Rules

f(0) → truef(1) → false
f(s(x)) → f(x)if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false

Strategy


The following SCCs where found

g#(x, c(y)) → g#(s(x), y)g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))
g#(x, c(y)) → g#(x, y)

f#(s(x)) → f#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

f#(s(x)) → f#(x)

Rewrite Rules

f(0) → truef(1) → false
f(s(x)) → f(x)if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

f#(s(x)) → f#(x)

Problem 3: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

g#(x, c(y)) → g#(s(x), y)g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))
g#(x, c(y)) → g#(x, y)

Rewrite Rules

f(0) → truef(1) → false
f(s(x)) → f(x)if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false

Strategy


Polynomial Interpretation

Improved Usable rules

if(false, s(x), s(y)) → s(y)if(true, s(x), s(y)) → s(x)

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

g#(x, c(y)) → g#(x, y)

Problem 4: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

g#(x, c(y)) → g#(s(x), y)g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))

Rewrite Rules

f(0) → truef(1) → false
f(s(x)) → f(x)if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, false, true

Strategy


Polynomial Interpretation

Improved Usable rules

if(false, s(x), s(y)) → s(y)if(true, s(x), s(y)) → s(x)
g(x, c(y)) → c(g(x, y))g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

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

g#(x, c(y)) → g#(s(x), y)

Problem 5: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))

Rewrite Rules

f(0) → truef(1) → false
f(s(x)) → f(x)if(true, s(x), s(y)) → s(x)
if(false, s(x), s(y)) → s(y)g(x, c(y)) → c(g(x, y))
g(x, c(y)) → g(x, if(f(x), c(g(s(x), y)), c(y)))

Original Signature

Termination of terms over the following signature is verified: f, g, 1, 0, s, c, if, true, false

Strategy


Polynomial Interpretation

Improved Usable rules

if(false, s(x), s(y)) → s(y)if(true, s(x), s(y)) → s(x)

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

g#(x, c(y)) → g#(x, if(f(x), c(g(s(x), y)), c(y)))