YES
 
The TRS could be proven terminating. The proof took 842 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (59ms).
 |  Problem 2 was processed with processor SubtermCriterion (1ms).
 |    |  Problem 5 was processed with processor DependencyGraph (0ms).
 |  Problem 3 was processed with processor PolynomialLinearRange4iUR (606ms).
 |    |  Problem 6 was processed with processor DependencyGraph (1ms).
 |  Problem 4 was processed with processor SubtermCriterion (1ms).
 Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| gcd#(s(x), s(y)) | → | le#(y, x) |  | if_minus#(false, s(x), y) | → | minus#(x, y) | 
| if_gcd#(false, s(x), s(y)) | → | gcd#(minus(y, x), s(x)) |  | le#(s(x), s(y)) | → | le#(x, y) | 
| minus#(s(x), y) | → | le#(s(x), y) |  | if_gcd#(true, s(x), s(y)) | → | minus#(x, y) | 
| minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) |  | if_gcd#(true, s(x), s(y)) | → | gcd#(minus(x, y), s(y)) | 
| gcd#(s(x), s(y)) | → | if_gcd#(le(y, x), s(x), s(y)) |  | if_gcd#(false, s(x), s(y)) | → | minus#(y, x) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, if_gcd, true, false, gcd
Strategy
The following SCCs where found
| if_minus#(false, s(x), y) → minus#(x, y) | minus#(s(x), y) → if_minus#(le(s(x), y), s(x), y) | 
| if_gcd#(false, s(x), s(y)) → gcd#(minus(y, x), s(x)) | if_gcd#(true, s(x), s(y)) → gcd#(minus(x, y), s(y)) | 
| gcd#(s(x), s(y)) → if_gcd#(le(y, x), s(x), s(y)) | 
| le#(s(x), s(y)) → le#(x, y) | 
 
 Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| if_minus#(false, s(x), y) | → | minus#(x, y) |  | minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, if_gcd, true, false, gcd
Strategy
Projection
The following projection was used:
- π (if_minus#): 2
- π (minus#): 1
Thus, the following dependency pairs are removed: 
| if_minus#(false, s(x), y) | → | minus#(x, y) | 
 
 Problem 5: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(s(x), y) | → | if_minus#(le(s(x), y), s(x), y) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, minus, 0, s, le, if_gcd, false, true, gcd
Strategy
There are no SCCs!
 
 Problem 3: PolynomialLinearRange4iUR
Dependency Pair Problem
Dependency Pairs
| if_gcd#(false, s(x), s(y)) | → | gcd#(minus(y, x), s(x)) |  | if_gcd#(true, s(x), s(y)) | → | gcd#(minus(x, y), s(y)) | 
| gcd#(s(x), s(y)) | → | if_gcd#(le(y, x), s(x), s(y)) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, if_gcd, true, false, gcd
Strategy
Polynomial Interpretation
- 0: 0
- false: 0
- gcd(x,y): 0
- gcd#(x,y): y + 2x
- if_gcd(x,y,z): 0
- if_gcd#(x,y,z): z + y + 1
- if_minus(x,y,z): y
- le(x,y): 2x
- minus(x,y): x
- s(x): 2x + 1
- true: 0
Improved Usable rules
| minus(0, y) | → | 0 |  | minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) | 
| if_minus(true, s(x), y) | → | 0 |  | if_minus(false, s(x), y) | → | s(minus(x, y)) | 
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| if_gcd#(false, s(x), s(y)) | → | gcd#(minus(y, x), s(x)) |  | if_gcd#(true, s(x), s(y)) | → | gcd#(minus(x, y), s(y)) | 
 
 Problem 6: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| gcd#(s(x), s(y)) | → | if_gcd#(le(y, x), s(x), s(y)) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, minus, 0, s, le, if_gcd, false, true, gcd
Strategy
There are no SCCs!
 
 Problem 4: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) | 
Rewrite Rules
| le(0, y) | → | true |  | le(s(x), 0) | → | false | 
| le(s(x), s(y)) | → | le(x, y) |  | minus(0, y) | → | 0 | 
| minus(s(x), y) | → | if_minus(le(s(x), y), s(x), y) |  | if_minus(true, s(x), y) | → | 0 | 
| if_minus(false, s(x), y) | → | s(minus(x, y)) |  | gcd(0, y) | → | y | 
| gcd(s(x), 0) | → | s(x) |  | gcd(s(x), s(y)) | → | if_gcd(le(y, x), s(x), s(y)) | 
| if_gcd(true, s(x), s(y)) | → | gcd(minus(x, y), s(y)) |  | if_gcd(false, s(x), s(y)) | → | gcd(minus(y, x), s(x)) | 
Original Signature
Termination of terms over the following signature is verified: if_minus, 0, minus, le, s, if_gcd, true, false, gcd
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed: 
| le#(s(x), s(y)) | → | le#(x, y) |