Certification Problem
Input (TPDB TRS_Standard/Der95/12)
The rewrite relation of the following TRS is considered.
not(not(x)) |
→ |
x |
(1) |
not(or(x,y)) |
→ |
and(not(x),not(y)) |
(2) |
not(and(x,y)) |
→ |
or(not(x),not(y)) |
(3) |
and(x,or(y,z)) |
→ |
or(and(x,y),and(x,z)) |
(4) |
and(or(y,z),x) |
→ |
or(and(x,y),and(x,z)) |
(5) |
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.
not#(or(x,y)) |
→ |
not#(x) |
(6) |
not#(and(x,y)) |
→ |
not#(x) |
(7) |
and#(x,or(y,z)) |
→ |
and#(x,z) |
(8) |
not#(and(x,y)) |
→ |
not#(y) |
(9) |
and#(or(y,z),x) |
→ |
and#(x,z) |
(10) |
not#(or(x,y)) |
→ |
not#(y) |
(11) |
not#(or(x,y)) |
→ |
and#(not(x),not(y)) |
(12) |
and#(x,or(y,z)) |
→ |
and#(x,y) |
(13) |
and#(or(y,z),x) |
→ |
and#(x,y) |
(14) |
1.1 Dependency Graph Processor
The dependency pairs are split into 2
components.