/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(x1) -> b(x1) b(a(c(x1))) -> c(c(a(a(x1)))) c(x1) -> b(x1) Proof: String Reversal Processor: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) DP Processor: DPs: c#(a(b(x1))) -> c#(x1) c#(a(b(x1))) -> c#(c(x1)) c#(a(b(x1))) -> a#(c(c(x1))) c#(a(b(x1))) -> a#(a(c(c(x1)))) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) TDG Processor: DPs: c#(a(b(x1))) -> c#(x1) c#(a(b(x1))) -> c#(c(x1)) c#(a(b(x1))) -> a#(c(c(x1))) c#(a(b(x1))) -> a#(a(c(c(x1)))) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) graph: c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> a#(a(c(c(x1)))) c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> a#(c(c(x1))) c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> c#(c(x1)) c#(a(b(x1))) -> c#(c(x1)) -> c#(a(b(x1))) -> c#(x1) c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> a#(a(c(c(x1)))) c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> a#(c(c(x1))) c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> c#(c(x1)) c#(a(b(x1))) -> c#(x1) -> c#(a(b(x1))) -> c#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 8/16 DPs: c#(a(b(x1))) -> c#(c(x1)) c#(a(b(x1))) -> c#(x1) TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) interpretation: [c#](x0) = [-& 3 ]x0 + [0], [-& 1 ] [0] [c](x0) = [-& 0 ]x0 + [0], [-& 1 ] [0] [b](x0) = [-& -&]x0 + [0], [0 1] [0] [a](x0) = [0 0]x0 + [1] orientation: c#(a(b(x1))) = [-& 4 ]x1 + [4] >= [-& 3 ]x1 + [3] = c#(c(x1)) c#(a(b(x1))) = [-& 4 ]x1 + [4] >= [-& 3 ]x1 + [0] = c#(x1) [0 1] [0] a(x1) = [0 0]x1 + [1] >= x1 = x1 [0 1] [0] [-& 1 ] [0] a(x1) = [0 0]x1 + [1] >= [-& -&]x1 + [0] = b(x1) [-& 2 ] [2] [-& 2 ] [2] c(a(b(x1))) = [-& 1 ]x1 + [1] >= [-& 1 ]x1 + [1] = a(a(c(c(x1)))) [-& 1 ] [0] [-& 1 ] [0] c(x1) = [-& 0 ]x1 + [0] >= [-& -&]x1 + [0] = b(x1) problem: DPs: TRS: a(x1) -> x1 a(x1) -> b(x1) c(a(b(x1))) -> a(a(c(c(x1)))) c(x1) -> b(x1) Qed