YES

The TRS could be proven terminating. The proof took 42 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (26ms).
 | – Problem 2 was processed with processor SubtermCriterion (3ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph



Dependency Pair Problem

Dependency Pairs

fib#(s(s(x)))sp#(g(x))g#(s(x))np#(g(x))
+#(x, s(y))+#(x, y)fib#(s(s(x)))g#(x)
np#(pair(x, y))+#(x, y)g#(s(x))g#(x)
sp#(pair(x, y))+#(x, y)

Rewrite Rules

fib(0)0fib(s(0))s(0)
fib(s(s(0)))s(0)fib(s(s(x)))sp(g(x))
g(0)pair(s(0), 0)g(s(0))pair(s(0), s(0))
g(s(x))np(g(x))sp(pair(x, y))+(x, y)
np(pair(x, y))pair(+(x, y), x)+(x, 0)x
+(x, s(y))s(+(x, y))

Original Signature

Termination of terms over the following signature is verified: g, 0, s, sp, pair, np, +, fib

Strategy


The following SCCs where found

+#(x, s(y)) → +#(x, y)

g#(s(x)) → g#(x)

Problem 2: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

g#(s(x))g#(x)

Rewrite Rules

fib(0)0fib(s(0))s(0)
fib(s(s(0)))s(0)fib(s(s(x)))sp(g(x))
g(0)pair(s(0), 0)g(s(0))pair(s(0), s(0))
g(s(x))np(g(x))sp(pair(x, y))+(x, y)
np(pair(x, y))pair(+(x, y), x)+(x, 0)x
+(x, s(y))s(+(x, y))

Original Signature

Termination of terms over the following signature is verified: g, 0, s, sp, pair, np, +, fib

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

g#(s(x))g#(x)

Problem 3: SubtermCriterion



Dependency Pair Problem

Dependency Pairs

+#(x, s(y))+#(x, y)

Rewrite Rules

fib(0)0fib(s(0))s(0)
fib(s(s(0)))s(0)fib(s(s(x)))sp(g(x))
g(0)pair(s(0), 0)g(s(0))pair(s(0), s(0))
g(s(x))np(g(x))sp(pair(x, y))+(x, y)
np(pair(x, y))pair(+(x, y), x)+(x, 0)x
+(x, s(y))s(+(x, y))

Original Signature

Termination of terms over the following signature is verified: g, 0, s, sp, pair, np, +, fib

Strategy


Projection

The following projection was used:

Thus, the following dependency pairs are removed:

+#(x, s(y))+#(x, y)