Certification Problem
Input (TPDB TRS_Standard/SK90/4.21)
The rewrite relation of the following TRS is considered.
|
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(1) |
|
and(x,and(y,y)) |
→ |
and(x,y) |
(2) |
|
or(or(x,y),and(y,z)) |
→ |
or(x,y) |
(3) |
|
or(x,and(x,y)) |
→ |
x |
(4) |
|
or(true,y) |
→ |
true |
(5) |
|
or(x,false) |
→ |
x |
(6) |
|
or(x,x) |
→ |
x |
(7) |
|
or(x,or(y,y)) |
→ |
or(x,y) |
(8) |
|
and(x,true) |
→ |
x |
(9) |
|
and(false,y) |
→ |
false |
(10) |
|
and(x,x) |
→ |
x |
(11) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by NaTT @ termCOMP 2023)
1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
and#(x,or(y,z)) |
→ |
or#(and(x,y),and(x,z)) |
(12) |
|
or#(x,or(y,y)) |
→ |
or#(x,y) |
(13) |
|
and#(x,or(y,z)) |
→ |
and#(x,y) |
(14) |
|
and#(x,and(y,y)) |
→ |
and#(x,y) |
(15) |
|
and#(x,or(y,z)) |
→ |
and#(x,z) |
(16) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.