YES

Problem 1: 

(VAR x y z)
(THEORY 
(AC plus))
(RULES
minus(minus(x,y),z) -> minus(x,plus(y,z))
minus(s(x),s(y)) -> minus(x,y)
minus(x,0) -> x
plus(0,y) -> y
plus(s(x),y) -> s(plus(x,y))
quot(0,s(y)) -> 0
quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
)

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 MINUS(minus(x,y),z) -> MINUS(x,plus(y,z))
 MINUS(minus(x,y),z) -> PLUS(y,z)
 MINUS(s(x),s(y)) -> MINUS(x,y)
 PLUS(plus(0,y),x3) -> PLUS(y,x3)
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
 QUOT(s(x),s(y)) -> MINUS(x,y)
 QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y))
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)

Problem 1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 MINUS(minus(x,y),z) -> MINUS(x,plus(y,z))
 MINUS(minus(x,y),z) -> PLUS(y,z)
 MINUS(s(x),s(y)) -> MINUS(x,y)
 PLUS(plus(0,y),x3) -> PLUS(y,x3)
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
 QUOT(s(x),s(y)) -> MINUS(x,y)
 QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y))
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 PLUS(plus(0,y),x3) -> PLUS(y,x3)
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
 PLUS(plus(x3,x4),x5) -> PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) -> PLUS(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->->Cycle:
->->-> Pairs:
 MINUS(minus(x,y),z) -> MINUS(x,plus(y,z))
 MINUS(s(x),s(y)) -> MINUS(x,y)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 Empty
->->Cycle:
->->-> Pairs:
 QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y))
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 Empty


The problem is decomposed in 3 subproblems.

Problem 1.1: 

Reduction Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(0,y),x3) -> PLUS(y,x3)
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 0
[plus](X1,X2) = X1 + X2
[quot](X1,X2) = 0
[0] = 2
[s](X) = X
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = X1 + X2
[QUOT](X1,X2) = 0

Problem 1.1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
 PLUS(plus(x3,x4),x5) -> PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) -> PLUS(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)

Problem 1.1: 

Reduction Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(plus(s(x),y),x3) -> PLUS(x,y)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 0
[plus](X1,X2) = X1 + X2 + 2
[quot](X1,X2) = 0
[0] = 0
[s](X) = X + 2
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = 2.X1 + 2.X2
[QUOT](X1,X2) = 0

Problem 1.1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
 PLUS(plus(x3,x4),x5) -> PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) -> PLUS(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)

Problem 1.1: 

Reduction Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 PLUS(plus(x3,x4),x5) -> PLUS(x3,x4)
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 0
[plus](X1,X2) = X1 + X2 + 1
[quot](X1,X2) = 0
[0] = 0
[s](X) = X + 2
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = 2.X1 + 2.X2
[QUOT](X1,X2) = 0

Problem 1.1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
 PLUS(plus(x3,x4),x5) -> PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) -> PLUS(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)

Problem 1.1: 

Reduction Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
 PLUS(s(x),y) -> PLUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 0
[plus](X1,X2) = X1 + X2 + 2
[quot](X1,X2) = 0
[0] = 0
[s](X) = X + 2
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = 2.X1 + 2.X2
[QUOT](X1,X2) = 0

Problem 1.1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
-> FAxioms:
 plus(plus(x3,x4),x5) -> plus(x3,plus(x4,x5))
 plus(x3,x4) -> plus(x4,x3)
 PLUS(plus(x3,x4),x5) -> PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) -> PLUS(x4,x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
->->-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)

Problem 1.1: 

Reduction Pairs Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 PLUS(plus(s(x),y),x3) -> PLUS(s(plus(x,y)),x3)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = 0
[plus](X1,X2) = X1 + X2 + 2
[quot](X1,X2) = 0
[0] = 0
[s](X) = 2
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = 2.X1 + 2.X2
[QUOT](X1,X2) = 0

Problem 1.1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x3,x4),x5) = PLUS(x3,plus(x4,x5))
 PLUS(x3,x4) = PLUS(x4,x3)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 PLUS(x3,plus(x4,x5)) -> PLUS(x4,x5)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> FAxioms:
 Empty
-> Pairs:
 MINUS(minus(x,y),z) -> MINUS(x,plus(y,z))
 MINUS(s(x),s(y)) -> MINUS(x,y)
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 Empty
->Projection:
 pi(MINUS) = [1]

Problem 1.2: 

SCC Processor:
-> FAxioms:
 Empty
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 Empty
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.3: 

Reduction Pairs Processor:
-> FAxioms:
 Empty
-> Pairs:
 QUOT(s(x),s(y)) -> QUOT(minus(x,y),s(y))
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Usable Equations:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> Usable Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
-> SRules:
 Empty
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[minus](X1,X2) = X1 + 1
[plus](X1,X2) = X1 + X2 + 2
[quot](X1,X2) = 0
[0] = 0
[s](X) = X + 2
[MINUS](X1,X2) = 0
[PLUS](X1,X2) = 0
[QUOT](X1,X2) = 2.X1

Problem 1.3: 

SCC Processor:
-> FAxioms:
 Empty
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x3,x4),x5) = plus(x3,plus(x4,x5))
 plus(x3,x4) = plus(x4,x3)
-> Rules:
 minus(minus(x,y),z) -> minus(x,plus(y,z))
 minus(s(x),s(y)) -> minus(x,y)
 minus(x,0) -> x
 plus(0,y) -> y
 plus(s(x),y) -> s(plus(x,y))
 quot(0,s(y)) -> 0
 quot(s(x),s(y)) -> s(quot(minus(x,y),s(y)))
-> SRules:
 Empty
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.