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