/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(a(x1)) -> b(x1) b(x1) -> x1 b(c(x1)) -> c(c(b(a(x1)))) c(x1) -> x1 Proof: String Reversal Processor: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 DP Processor: DPs: a#(a(x1)) -> b#(x1) c#(b(x1)) -> c#(x1) c#(b(x1)) -> c#(c(x1)) c#(b(x1)) -> b#(c(c(x1))) c#(b(x1)) -> a#(b(c(c(x1)))) TRS: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 TDG Processor: DPs: a#(a(x1)) -> b#(x1) c#(b(x1)) -> c#(x1) c#(b(x1)) -> c#(c(x1)) c#(b(x1)) -> b#(c(c(x1))) c#(b(x1)) -> a#(b(c(c(x1)))) TRS: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 graph: c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> a#(b(c(c(x1)))) c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> b#(c(c(x1))) c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> c#(c(x1)) c#(b(x1)) -> c#(c(x1)) -> c#(b(x1)) -> c#(x1) c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> a#(b(c(c(x1)))) c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> b#(c(c(x1))) c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> c#(c(x1)) c#(b(x1)) -> c#(x1) -> c#(b(x1)) -> c#(x1) c#(b(x1)) -> a#(b(c(c(x1)))) -> a#(a(x1)) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 9/25 DPs: c#(b(x1)) -> c#(c(x1)) c#(b(x1)) -> c#(x1) TRS: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 interpretation: [c#](x0) = [-& 0 ]x0 + [0], [0 0] [0] [c](x0) = [0 0]x0 + [2], [0 0] [-&] [b](x0) = [1 1]x0 + [3 ], [0 0] [3 ] [a](x0) = [1 0]x0 + [-&] orientation: c#(b(x1)) = [1 1]x1 + [3] >= [0 0]x1 + [2] = c#(c(x1)) c#(b(x1)) = [1 1]x1 + [3] >= [-& 0 ]x1 + [0] = c#(x1) [0 0] [3 ] a(x1) = [1 0]x1 + [-&] >= x1 = x1 [1 0] [3] [0 0] [-&] a(a(x1)) = [1 1]x1 + [4] >= [1 1]x1 + [3 ] = b(x1) [0 0] [-&] b(x1) = [1 1]x1 + [3 ] >= x1 = x1 [1 1] [3] [1 1] [3] c(b(x1)) = [1 1]x1 + [3] >= [1 1]x1 + [3] = a(b(c(c(x1)))) [0 0] [0] c(x1) = [0 0]x1 + [2] >= x1 = x1 problem: DPs: TRS: a(x1) -> x1 a(a(x1)) -> b(x1) b(x1) -> x1 c(b(x1)) -> a(b(c(c(x1)))) c(x1) -> x1 Qed