R
↳Dependency Pair Analysis
P(p(s(x))) -> P(x)
LE(p(s(x)), x) -> LE(x, x)
LE(s(x), s(y)) -> LE(x, y)
MINUS(x, y) -> IF(le(x, y), x, y)
MINUS(x, y) -> LE(x, y)
IF(false, x, y) -> MINUS(p(x), y)
IF(false, x, y) -> P(x)
R
↳DPs
→DP Problem 1
↳Argument Filtering and Ordering
→DP Problem 2
↳Remaining
LE(s(x), s(y)) -> LE(x, y)
p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
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)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))
innermost
LE(s(x), s(y)) -> LE(x, y)
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 3
↳Dependency Graph
→DP Problem 2
↳Remaining
p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
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)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))
innermost
R
↳DPs
→DP Problem 1
↳AFS
→DP Problem 2
↳Remaining Obligation(s)
IF(false, x, y) -> MINUS(p(x), y)
MINUS(x, y) -> IF(le(x, y), x, y)
p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
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)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))
innermost