Certification Problem
Input (TPDB TRS_Standard/SK90/2.44)
The rewrite relation of the following TRS is considered.
|
del(.(x,.(y,z))) |
→ |
f(=(x,y),x,y,z) |
(1) |
|
f(true,x,y,z) |
→ |
del(.(y,z)) |
(2) |
|
f(false,x,y,z) |
→ |
.(x,del(.(y,z))) |
(3) |
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
Property / Task
Prove or disprove termination.Answer / Result
Yes.Proof (by AProVE @ termCOMP 2023)
1 Switch to Innermost Termination
The TRS is overlay and locally confluent:
10Hence, it suffices to show innermost termination in the following.
1.1 Dependency Pair Transformation
The following set of initial dependency pairs has been identified.
|
del#(.(x,.(y,z))) |
→ |
f#(=(x,y),x,y,z) |
(8) |
|
del#(.(x,.(y,z))) |
→ |
=#(x,y) |
(9) |
|
f#(true,x,y,z) |
→ |
del#(.(y,z)) |
(10) |
|
f#(false,x,y,z) |
→ |
del#(.(y,z)) |
(11) |
|
=#(.(x,y),.(u,v)) |
→ |
=#(x,u) |
(12) |
|
=#(.(x,y),.(u,v)) |
→ |
=#(y,v) |
(13) |
1.1.1 Dependency Graph Processor
The dependency pairs are split into 1
component.
-
The
1st
component contains the
pair
|
f#(true,x,y,z) |
→ |
del#(.(y,z)) |
(10) |
|
del#(.(x,.(y,z))) |
→ |
f#(=(x,y),x,y,z) |
(8) |
|
f#(false,x,y,z) |
→ |
del#(.(y,z)) |
(11) |
1.1.1.1 Usable Rules Processor
We restrict the rewrite rules to the following usable rules of the DP problem.
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
1.1.1.1.1 Innermost Lhss Removal Processor
We restrict the innermost strategy to the following left hand sides.
|
=(nil,nil) |
|
=(.(x0,x1),nil) |
|
=(nil,.(x0,x1)) |
|
=(.(x0,x1),.(u,v)) |
1.1.1.1.1.1 Monotonic Reduction Pair Processor with Usable Rules
Using the linear polynomial interpretation over the naturals
| [=(x1, x2)] |
= |
1 · x1 + 2 · x2
|
| [nil] |
= |
1 |
| [true] |
= |
1 |
| [.(x1, x2)] |
= |
2 + 2 · x1 + 2 · x2
|
| [false] |
= |
1 |
| [u] |
= |
2 |
| [v] |
= |
0 |
| [and(x1, x2)] |
= |
1 · x1 + 1 · x2
|
| [f#(x1,...,x4)] |
= |
1 + 1 · x1 + 1 · x2 + 2 · x3 + 2 · x4
|
| [del#(x1)] |
= |
1 · x1
|
together with the usable
rules
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
(w.r.t. the implicit argument filter of the reduction pair),
the
pair
|
del#(.(x,.(y,z))) |
→ |
f#(=(x,y),x,y,z) |
(8) |
and
the
rules
|
=(nil,nil) |
→ |
true |
(4) |
|
=(.(x,y),nil) |
→ |
false |
(5) |
|
=(nil,.(y,z)) |
→ |
false |
(6) |
|
=(.(x,y),.(u,v)) |
→ |
and(=(x,u),=(y,v)) |
(7) |
could be deleted.
1.1.1.1.1.1.1 Dependency Graph Processor
The dependency pairs are split into 0
components.