/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: b(b(a(a(x1)))) -> b(b(a(b(x1)))) b(a(b(b(x1)))) -> a(a(a(a(x1)))) b(a(b(b(x1)))) -> b(b(a(a(x1)))) Proof: Matrix Interpretation Processor: dim=4 interpretation: [1 0 0 1] [0] [0 0 0 0] [0] [b](x0) = [0 0 0 0]x0 + [0] [0 1 0 0] [1], [1 0 0 0] [0 0 1 0] [a](x0) = [0 0 0 1]x0 [0 0 0 0] orientation: [1 0 0 1] [1] [1 0 0 1] [1] [0 0 0 0] [0] [0 0 0 0] [0] b(b(a(a(x1)))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = b(b(a(b(x1)))) [0 0 0 0] [1] [0 0 0 0] [1] [1 1 0 1] [1] [1 0 0 0] [0 0 0 0] [0] [0 0 0 0] b(a(b(b(x1)))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 = a(a(a(a(x1)))) [0 0 0 0] [1] [0 0 0 0] [1 1 0 1] [1] [1 0 0 1] [1] [0 0 0 0] [0] [0 0 0 0] [0] b(a(b(b(x1)))) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = b(b(a(a(x1)))) [0 0 0 0] [1] [0 0 0 0] [1] problem: b(b(a(a(x1)))) -> b(b(a(b(x1)))) b(a(b(b(x1)))) -> b(b(a(a(x1)))) KBO Processor: weight function: w0 = 1 w(b) = w(a) = 1 precedence: a > b problem: Qed