/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


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YES

Problem 1: 

(VAR x y)
(RULES
if(false,x,y) -> s(minus(p(x),y))
if(true,x,y) -> 0
le(0,y) -> true
le(s(x),0) -> false
le(s(x),s(y)) -> le(x,y)
minus(x,y) -> if(le(x,y),x,y)
p(0) -> 0
p(s(x)) -> x
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 IF(false,x,y) -> MINUS(p(x),y)
 IF(false,x,y) -> P(x)
 LE(s(x),s(y)) -> LE(x,y)
 MINUS(x,y) -> IF(le(x,y),x,y)
 MINUS(x,y) -> LE(x,y)
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x

Problem 1: 

SCC Processor:
-> Pairs:
 IF(false,x,y) -> MINUS(p(x),y)
 IF(false,x,y) -> P(x)
 LE(s(x),s(y)) -> LE(x,y)
 MINUS(x,y) -> IF(le(x,y),x,y)
 MINUS(x,y) -> LE(x,y)
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 LE(s(x),s(y)) -> LE(x,y)
->->-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
->->Cycle:
->->-> Pairs:
 IF(false,x,y) -> MINUS(p(x),y)
 MINUS(x,y) -> IF(le(x,y),x,y)
->->-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 LE(s(x),s(y)) -> LE(x,y)
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
->Projection:
 pi(LE) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Reduction Pairs Processor:
-> Pairs:
 IF(false,x,y) -> MINUS(p(x),y)
 MINUS(x,y) -> IF(le(x,y),x,y)
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
-> Usable rules:
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 p(0) -> 0
 p(s(x)) -> x
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[le](X1,X2) = 1/2.X1
[p](X) = 1/2.X
[0] = 0
[false] = 1/2
[s](X) = 2.X + 2
[true] = 0
[IF](X1,X2,X3) = X1 + 1/2.X2 + 1/2.X3 + 1/2
[MINUS](X1,X2) = X1 + 1/2.X2 + 1/2

Problem 1.2: 

SCC Processor:
-> Pairs:
 MINUS(x,y) -> IF(le(x,y),x,y)
-> Rules:
 if(false,x,y) -> s(minus(p(x),y))
 if(true,x,y) -> 0
 le(0,y) -> true
 le(s(x),0) -> false
 le(s(x),s(y)) -> le(x,y)
 minus(x,y) -> if(le(x,y),x,y)
 p(0) -> 0
 p(s(x)) -> x
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.