Term Rewriting System R:
[x, y]
not(x) -> if(x, false, true)
and(x, y) -> if(x, y, false)
or(x, y) -> if(x, true, y)
implies(x, y) -> if(x, y, true)
=(x, x) -> true
=(x, y) -> if(x, y, not(y))
=(x, y) -> if(x, y, if(y, false, true))
if(true, x, y) -> x
if(false, x, y) -> y
if(x, x, if(x, false, true)) -> true
Termination of R to be shown.
R
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
not(x) -> if(x, false, true)
=(x, x) -> true
=(x, y) -> if(x, y, if(y, false, true))
where the Polynomial interpretation:
| POL(and(x1, x2)) | = x1 + x2 |
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(false) | = 0 |
| POL(=(x1, x2)) | = 1 + x1 + 2·x2 |
| POL(or(x1, x2)) | = x1 + x2 |
| POL(true) | = 0 |
| POL(implies(x1, x2)) | = x1 + x2 |
| POL(not(x1)) | = 1 + x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
or(x, y) -> if(x, true, y)
where the Polynomial interpretation:
| POL(and(x1, x2)) | = x1 + x2 |
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(false) | = 0 |
| POL(=(x1, x2)) | = x1 + 2·x2 |
| POL(true) | = 0 |
| POL(implies(x1, x2)) | = x1 + x2 |
| POL(or(x1, x2)) | = 1 + x1 + x2 |
| POL(not(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
if(true, x, y) -> x
where the Polynomial interpretation:
| POL(and(x1, x2)) | = x1 + x2 |
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(false) | = 0 |
| POL(=(x1, x2)) | = x1 + 2·x2 |
| POL(true) | = 1 |
| POL(implies(x1, x2)) | = 1 + x1 + x2 |
| POL(not(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS4
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
if(false, x, y) -> y
if(x, x, if(x, false, true)) -> true
where the Polynomial interpretation:
| POL(and(x1, x2)) | = 1 + x1 + x2 |
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(false) | = 1 |
| POL(=(x1, x2)) | = x1 + 2·x2 |
| POL(true) | = 0 |
| POL(implies(x1, x2)) | = x1 + x2 |
| POL(not(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS5
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
and(x, y) -> if(x, y, false)
where the Polynomial interpretation:
| POL(and(x1, x2)) | = 1 + x1 + x2 |
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(false) | = 0 |
| POL(=(x1, x2)) | = x1 + 2·x2 |
| POL(implies(x1, x2)) | = x1 + x2 |
| POL(true) | = 0 |
| POL(not(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS6
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
=(x, y) -> if(x, y, not(y))
where the Polynomial interpretation:
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(=(x1, x2)) | = 1 + x1 + 2·x2 |
| POL(implies(x1, x2)) | = x1 + x2 |
| POL(true) | = 0 |
| POL(not(x1)) | = x1 |
was used.
Not all Rules of R can be deleted, so we still have to regard a part of R.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS7
↳Removing Redundant Rules
Removing the following rules from R which fullfill a polynomial ordering:
implies(x, y) -> if(x, y, true)
where the Polynomial interpretation:
| POL(if(x1, x2, x3)) | = x1 + x2 + x3 |
| POL(implies(x1, x2)) | = 1 + x1 + x2 |
| POL(true) | = 0 |
was used.
All Rules of R can be deleted.
R
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS8
↳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
↳RRRPolo
→TRS2
↳RRRPolo
→TRS3
↳RRRPolo
...
→TRS9
↳Dependency Pair Analysis
R contains no Dependency Pairs and therefore no SCCs.
Termination of R successfully shown.
Duration:
0:00 minutes