Term Rewriting System R:
[y, x, xs, f]
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
Termination of R to be shown.
R
↳Overlay and local confluence Check
The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.
R
↳OC
→TRS2
↳Dependency Pair Analysis
R contains the following Dependency Pairs:
APP(app(plus, app(s, x)), y) -> APP(s, app(app(plus, x), y))
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
APP(app(plus, app(s, x)), y) -> APP(plus, x)
APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
APP(inc, xs) -> APP(map, app(plus, app(s, 0)))
APP(inc, xs) -> APP(plus, app(s, 0))
APP(inc, xs) -> APP(s, 0)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(cons, app(f, x)), app(app(map, f), xs))
APP(app(map, f), app(app(cons, x), xs)) -> APP(cons, app(f, x))
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
Furthermore, R contains two SCCs.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳Usable Rules (Innermost)
→DP Problem 2
↳UsableRules
Dependency Pair:
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
Rules:
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
Strategy:
innermost
As we are in the innermost case, we can delete all 5 non-usable-rules.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
...
→DP Problem 3
↳A-Transformation
→DP Problem 2
↳UsableRules
Dependency Pair:
APP(app(plus, app(s, x)), y) -> APP(app(plus, x), y)
Rule:
none
Strategy:
innermost
We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
...
→DP Problem 4
↳Size-Change Principle
→DP Problem 2
↳UsableRules
Dependency Pair:
PLUS(s(x), y) -> PLUS(x, y)
Rule:
none
Strategy:
innermost
We number the DPs as follows:
- PLUS(s(x), y) -> PLUS(x, y)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
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.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳Usable Rules (Innermost)
Dependency Pairs:
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
Rules:
app(app(plus, 0), y) -> y
app(app(plus, app(s, x)), y) -> app(s, app(app(plus, x), y))
app(inc, xs) -> app(app(map, app(plus, app(s, 0))), xs)
app(app(map, f), nil) -> nil
app(app(map, f), app(app(cons, x), xs)) -> app(app(cons, app(f, x)), app(app(map, f), xs))
Strategy:
innermost
As we are in the innermost case, we can delete all 5 non-usable-rules.
R
↳OC
→TRS2
↳DPs
→DP Problem 1
↳UsableRules
→DP Problem 2
↳UsableRules
...
→DP Problem 5
↳Size-Change Principle
Dependency Pairs:
APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
Rule:
none
Strategy:
innermost
We number the DPs as follows:
- APP(app(map, f), app(app(cons, x), xs)) -> APP(app(map, f), xs)
- APP(app(map, f), app(app(cons, x), xs)) -> APP(f, x)
- APP(inc, xs) -> APP(app(map, app(plus, app(s, 0))), xs)
and get the following Size-Change Graph(s):
which lead(s) to this/these maximal multigraph(s):
DP: empty set
Oriented Rules: none
We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial
with Argument Filtering System:
app(x1, x2) -> app(x1, x2)
We obtain no new DP problems.
Termination of R successfully shown.
Duration:
0:00 minutes