YES
The TRS could be proven terminating. The proof took 36 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (20ms).
| Problem 2 was processed with processor SubtermCriterion (1ms).
| Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
and#(x, or(y, z)) | → | and#(x, z) | | not#(and(x, y)) | → | not#(x) |
not#(or(x, y)) | → | and#(not(x), not(y)) | | not#(or(x, y)) | → | not#(x) |
and#(x, or(y, z)) | → | and#(x, y) | | not#(and(x, y)) | → | not#(y) |
not#(or(x, y)) | → | not#(y) |
Rewrite Rules
not(and(x, y)) | → | or(not(x), not(y)) | | not(or(x, y)) | → | and(not(x), not(y)) |
and(x, or(y, z)) | → | or(and(x, y), and(x, z)) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
The following SCCs where found
not#(and(x, y)) → not#(x) | not#(or(x, y)) → not#(x) |
not#(and(x, y)) → not#(y) | not#(or(x, y)) → not#(y) |
and#(x, or(y, z)) → and#(x, z) | and#(x, or(y, z)) → and#(x, y) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
and#(x, or(y, z)) | → | and#(x, z) | | and#(x, or(y, z)) | → | and#(x, y) |
Rewrite Rules
not(and(x, y)) | → | or(not(x), not(y)) | | not(or(x, y)) | → | and(not(x), not(y)) |
and(x, or(y, z)) | → | or(and(x, y), and(x, z)) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
and#(x, or(y, z)) | → | and#(x, z) | | and#(x, or(y, z)) | → | and#(x, y) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
not#(and(x, y)) | → | not#(x) | | not#(or(x, y)) | → | not#(x) |
not#(and(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(y) |
Rewrite Rules
not(and(x, y)) | → | or(not(x), not(y)) | | not(or(x, y)) | → | and(not(x), not(y)) |
and(x, or(y, z)) | → | or(and(x, y), and(x, z)) |
Original Signature
Termination of terms over the following signature is verified: not, or, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
not#(and(x, y)) | → | not#(x) | | not#(or(x, y)) | → | not#(x) |
not#(and(x, y)) | → | not#(y) | | not#(or(x, y)) | → | not#(y) |