Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MSORT(.(x, y)) -> MIN(x, y)
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))
MSORT(.(x, y)) -> DEL(min(x, y), .(x, y))
MIN(x, .(y, z)) -> MIN(x, z)
MIN(x, .(y, z)) -> MIN(y, z)
DEL(x, .(y, z)) -> DEL(x, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MIN(x, .(y, z)) -> MIN(y, z)
MIN(x, .(y, z)) -> MIN(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
DEL(x, .(y, z)) -> DEL(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MIN(x, .(y, z)) -> MIN(y, z)
MIN(x, .(y, z)) -> MIN(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
DEL(x, .(y, z)) -> DEL(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

     ↳DPs

       →DP Problem 1

         ↳Remaining Obligation(s)

       →DP Problem 2

         ↳Remaining Obligation(s)

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
MIN(x, .(y, z)) -> MIN(y, z)
MIN(x, .(y, z)) -> MIN(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
DEL(x, .(y, z)) -> DEL(x, z)

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))
Dependency Pair:
MSORT(.(x, y)) -> MSORT(del(min(x, y), .(x, y)))

Rules:

msort(nil) -> nil
msort(.(x, y)) -> .(min(x, y), msort(del(min(x, y), .(x, y))))
min(x, nil) -> x
min(x, .(y, z)) -> if(<=(x, y), min(x, z), min(y, z))
del(x, nil) -> nil
del(x, .(y, z)) -> if(=(x, y), z, .(y, del(x, z)))

Termination of R could not be shown.
Duration:
0:00 minutes