Term Rewriting System R:
[y, z, x]
quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))

Furthermore, R contains one SCC.

R
DPs
→DP Problem 1
Usable Rules (Innermost)

Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, s(z), s(z)))

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

R
DPs
→DP Problem 1
UsableRules
→DP Problem 2
Size-Change Principle

Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. QUOT(x, 0, s(z)) -> QUOT(x, s(z), s(z))
2. QUOT(s(x), s(y), z) -> QUOT(x, y, z)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
3=2
3=3
{2} , {2}
1>1
2>2
3=3

which lead(s) to this/these maximal multigraph(s):
{2} , {2}
1>1
2>2
3=3
{1} , {2}
1>1
3>2
3=3
{2} , {1}
1>1
3=2
3=3
{2} , {2}
1>1
3>2
3=3

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes