R
↳Dependency Pair Analysis
MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) -> MINUS(s(x), s(y))
R
↳DPs
→DP Problem 1
↳Argument Filtering and Ordering
→DP Problem 2
↳AFS
→DP Problem 3
↳Rw
MINUS(s(x), s(y)) -> MINUS(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
MINUS(s(x), s(y)) -> MINUS(x, y)
POL(MINUS(x1, x2)) = x1 + x2 POL(s(x1)) = 1 + x1
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 4
↳Dependency Graph
→DP Problem 2
↳AFS
→DP Problem 3
↳Rw
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳Argument Filtering and Ordering
→DP Problem 3
↳Rw
LE(s(x), s(y)) -> LE(x, y)
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
LE(s(x), s(y)) -> LE(x, y)
POL(LE(x1, x2)) = x1 + x2 POL(s(x1)) = 1 + x1
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 5
↳Dependency Graph
→DP Problem 3
↳Rw
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 3
↳Rewriting Transformation
QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
one new Dependency Pair is created:
QUOT(s(x), s(y)) -> QUOT(minus(s(x), s(y)), s(y))
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 3
↳Rw
→DP Problem 6
↳Argument Filtering and Ordering
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
POL(QUOT(x1, x2)) = x1 + x2 POL(s(x1)) = 1 + x1
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
minus(x1, x2) -> x1
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳AFS
→DP Problem 3
↳Rw
→DP Problem 6
↳AFS
...
→DP Problem 7
↳Dependency Graph
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(s(x), s(y)), s(y)))
innermost