YES
The TRS could be proven terminating. The proof took 60000 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (44ms).
| – Problem 2 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| or#(x, y) | → | xor#(x, y) | | implies#(x, y) | → | xor#(and(x, y), xor(x, true)) |
| implies#(x, y) | → | xor#(x, true) | | and#(xor(x, y), z) | → | xor#(and(x, z), and(y, z)) |
| or#(x, y) | → | and#(x, y) | | and#(xor(x, y), z) | → | and#(y, z) |
| implies#(x, y) | → | and#(x, y) | | and#(xor(x, y), z) | → | and#(x, z) |
| or#(x, y) | → | xor#(and(x, y), xor(x, y)) | | not#(x) | → | xor#(x, true) |
Rewrite Rules
| not(x) | → | xor(x, true) | | or(x, y) | → | xor(and(x, y), xor(x, y)) |
| implies(x, y) | → | xor(and(x, y), xor(x, true)) | | and(x, true) | → | x |
| and(x, false) | → | false | | and(x, x) | → | x |
| xor(x, false) | → | x | | xor(x, x) | → | false |
| and(xor(x, y), z) | → | xor(and(x, z), and(y, z)) |
Original Signature
Termination of terms over the following signature is verified: not, xor, or, implies, true, false, and
Strategy
The following SCCs where found
| and#(xor(x, y), z) → and#(y, z) | and#(xor(x, y), z) → and#(x, z) |
Problem 2: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| and#(xor(x, y), z) | → | and#(y, z) | | and#(xor(x, y), z) | → | and#(x, z) |
Rewrite Rules
| not(x) | → | xor(x, true) | | or(x, y) | → | xor(and(x, y), xor(x, y)) |
| implies(x, y) | → | xor(and(x, y), xor(x, true)) | | and(x, true) | → | x |
| and(x, false) | → | false | | and(x, x) | → | x |
| xor(x, false) | → | x | | xor(x, x) | → | false |
| and(xor(x, y), z) | → | xor(and(x, z), and(y, z)) |
Original Signature
Termination of terms over the following signature is verified: not, xor, or, implies, true, false, and
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| and#(xor(x, y), z) | → | and#(y, z) | | and#(xor(x, y), z) | → | and#(x, z) |