R
↳Dependency Pair Analysis
MINUS(+(x, y)) -> +'(minus(y), minus(x))
MINUS(+(x, y)) -> MINUS(y)
MINUS(+(x, y)) -> MINUS(x)
*'(x, +(y, z)) -> +'(*(x, y), *(x, z))
*'(x, +(y, z)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, minus(y)) -> MINUS(*(x, y))
*'(x, minus(y)) -> *'(x, y)
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↳DPs
→DP Problem 1
↳Argument Filtering and Ordering
→DP Problem 2
↳AFS
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(y)
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))
MINUS(+(x, y)) -> MINUS(x)
MINUS(+(x, y)) -> MINUS(y)
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))
* > minus > {+, 0}
MINUS(x1) -> MINUS(x1)
+(x1, x2) -> +(x1, x2)
minus(x1) -> minus(x1)
*(x1, x2) -> *(x1, x2)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳AFS
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳Argument Filtering and Ordering
*'(x, minus(y)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))
*'(x, minus(y)) -> *'(x, y)
*'(x, +(y, z)) -> *'(x, z)
*'(x, +(y, z)) -> *'(x, y)
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))
* > {0, minus} > +
*'(x1, x2) -> *'(x1, x2)
+(x1, x2) -> +(x1, x2)
minus(x1) -> minus(x1)
*(x1, x2) -> *(x1, x2)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 4
↳Dependency Graph
+(x, 0) -> x
+(minus(x), x) -> 0
minus(0) -> 0
minus(minus(x)) -> x
minus(+(x, y)) -> +(minus(y), minus(x))
*(x, 1) -> x
*(x, 0) -> 0
*(x, +(y, z)) -> +(*(x, y), *(x, z))
*(x, minus(y)) -> minus(*(x, y))