/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES Problem: a(x1) -> x1 a(b(x1)) -> x1 b(a(x1)) -> a(a(b(c(x1)))) c(c(x1)) -> b(x1) Proof: String Reversal Processor: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) DP Processor: DPs: a#(b(x1)) -> a#(x1) a#(b(x1)) -> a#(a(x1)) a#(b(x1)) -> b#(a(a(x1))) a#(b(x1)) -> c#(b(a(a(x1)))) c#(c(x1)) -> b#(x1) TRS: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) TDG Processor: DPs: a#(b(x1)) -> a#(x1) a#(b(x1)) -> a#(a(x1)) a#(b(x1)) -> b#(a(a(x1))) a#(b(x1)) -> c#(b(a(a(x1)))) c#(c(x1)) -> b#(x1) TRS: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) graph: a#(b(x1)) -> c#(b(a(a(x1)))) -> c#(c(x1)) -> b#(x1) a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> c#(b(a(a(x1)))) a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> b#(a(a(x1))) a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(a(x1)) a#(b(x1)) -> a#(a(x1)) -> a#(b(x1)) -> a#(x1) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> c#(b(a(a(x1)))) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> b#(a(a(x1))) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(a(x1)) a#(b(x1)) -> a#(x1) -> a#(b(x1)) -> a#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 9/25 DPs: a#(b(x1)) -> a#(a(x1)) a#(b(x1)) -> a#(x1) TRS: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) interpretation: [a#](x0) = [0 0]x0 + [0], [-& 1 ] [0] [c](x0) = [0 1 ]x0 + [3], [1 1] [3 ] [b](x0) = [0 0]x0 + [-&], [0 0] [-&] [a](x0) = [0 0]x0 + [2 ] orientation: a#(b(x1)) = [1 1]x1 + [3] >= [0 0]x1 + [2] = a#(a(x1)) a#(b(x1)) = [1 1]x1 + [3] >= [0 0]x1 + [0] = a#(x1) [0 0] [-&] a(x1) = [0 0]x1 + [2 ] >= x1 = x1 [1 1] [3] b(a(x1)) = [0 0]x1 + [2] >= x1 = x1 [1 1] [3] [1 1] [3] a(b(x1)) = [1 1]x1 + [3] >= [1 1]x1 + [3] = c(b(a(a(x1)))) [1 2] [4] [1 1] [3 ] c(c(x1)) = [1 2]x1 + [4] >= [0 0]x1 + [-&] = b(x1) problem: DPs: TRS: a(x1) -> x1 b(a(x1)) -> x1 a(b(x1)) -> c(b(a(a(x1)))) c(c(x1)) -> b(x1) Qed