/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. !870 : [o * o] --> o and : [o * o] --> o implies : [o * o] --> o not : [o] --> o or : [o * o] --> o true : [] --> o xor : [o * o] --> o not(X) => xor(X, true) implies(X, Y) => xor(and(X, Y), xor(X, true)) or(X, Y) => xor(and(X, Y), xor(X, Y)) !870(X, Y) => xor(X, xor(Y, true)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): not(X) >? xor(X, true) implies(X, Y) >? xor(and(X, Y), xor(X, true)) or(X, Y) >? xor(and(X, Y), xor(X, Y)) !870(X, Y) >? xor(X, xor(Y, true)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: !870 = \y0y1.3 + y0 + y1 and = \y0y1.y0 + y1 implies = \y0y1.3 + y1 + 3y0 not = \y0.3 + y0 or = \y0y1.3 + 3y0 + 3y1 true = 0 xor = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[not(_x0)]] = 3 + x0 > x0 = [[xor(_x0, true)]] [[implies(_x0, _x1)]] = 3 + x1 + 3x0 > x1 + 2x0 = [[xor(and(_x0, _x1), xor(_x0, true))]] [[or(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2x0 + 2x1 = [[xor(and(_x0, _x1), xor(_x0, _x1))]] [[!870(_x0, _x1)]] = 3 + x0 + x1 > x0 + x1 = [[xor(_x0, xor(_x1, true))]] We can thus remove the following rules: not(X) => xor(X, true) implies(X, Y) => xor(and(X, Y), xor(X, true)) or(X, Y) => xor(and(X, Y), xor(X, Y)) !870(X, Y) => xor(X, xor(Y, true)) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.