YES

The TRS could be proven terminating. The proof took 24 ms.

The following DP Processors were used


Problem 1 was processed with processor DependencyGraph (11ms).
 | – Problem 2 was processed with processor SubtermCriterion (1ms).
 | – Problem 3 was processed with processor SubtermCriterion (1ms).

Problem 1: DependencyGraph

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

fib(0) → 0fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Original Signature

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

Strategy

The following SCCs where found

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

Problem 2: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

+#(x, s(y)) → +#(x, y)

Rewrite Rules

fib(0) → 0fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Original Signature

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

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

+#(x, s(y)) → +#(x, y)

Problem 3: SubtermCriterion

Dependency Pair Problem

Dependency Pairs

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

Rewrite Rules

fib(0) → 0fib(s(0)) → s(0)
fib(s(s(x))) → +(fib(s(x)), fib(x))+(x, 0) → x
+(x, s(y)) → s(+(x, y))

Original Signature

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

Strategy

Projection

The following projection was used:

Thus, the following dependency pairs are removed:

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