YES

The TRS could be proven terminating. The proof took 726 ms.

The following DP Processors were used


Problem 1 was processed with processor PolynomialLinearRange4iUR (480ms).
 | – Problem 2 was processed with processor PolynomialLinearRange4iUR (78ms).

Problem 1: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

f#(x, y)ack#(x, y)f#(x, s(y))f#(y, x)
ack#(s(x), y)f#(x, x)ack#(s(x), s(y))ack#(x, ack(s(x), y))
ack#(s(x), s(y))ack#(s(x), y)ack#(s(x), 0)ack#(x, s(0))
f#(s(x), y)f#(x, s(x))

Rewrite Rules

ack(0, y)s(y)ack(s(x), 0)ack(x, s(0))
ack(s(x), s(y))ack(x, ack(s(x), y))f(s(x), y)f(x, s(x))
f(x, s(y))f(y, x)f(x, y)ack(x, y)
ack(s(x), y)f(x, x)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, ack

Strategy


Polynomial Interpretation

Improved Usable rules

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

f#(x, y)ack#(x, y)ack#(s(x), y)f#(x, x)
f#(x, s(y))f#(y, x)ack#(s(x), s(y))ack#(x, ack(s(x), y))
ack#(s(x), 0)ack#(x, s(0))f#(s(x), y)f#(x, s(x))

Problem 2: PolynomialLinearRange4iUR



Dependency Pair Problem

Dependency Pairs

ack#(s(x), s(y))ack#(s(x), y)

Rewrite Rules

ack(0, y)s(y)ack(s(x), 0)ack(x, s(0))
ack(s(x), s(y))ack(x, ack(s(x), y))f(s(x), y)f(x, s(x))
f(x, s(y))f(y, x)f(x, y)ack(x, y)
ack(s(x), y)f(x, x)

Original Signature

Termination of terms over the following signature is verified: f, 0, s, ack

Strategy


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:

ack#(s(x), s(y))ack#(s(x), y)