Termination Proof using AProVE

Term Rewriting System R:
[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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0

Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
IF_QUOT(true, x, y) -> MINUS(x, y)

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 Pair:
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pair:
LE(s(x), s(y)) -> LE(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pairs:
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0

     ↳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 Pair:
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pair:
LE(s(x), s(y)) -> LE(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pairs:
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0

     ↳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 Pair:
MINUS(s(x), s(y)) -> MINUS(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pair:
LE(s(x), s(y)) -> LE(x, y)

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0
Dependency Pairs:
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
if_quot(true, x, y) -> s(quot(minus(x, y), y))
if_quot(false, x, y) -> 0

Termination of R could not be shown.
Duration:
0:23 minutes