/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES Problem: a(x1) -> x1 a(b(x1)) -> b(b(c(a(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 Proof: String Reversal Processor: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 DP Processor: DPs: b#(a(x1)) -> b#(x1) b#(a(x1)) -> b#(b(x1)) b#(a(x1)) -> c#(b(b(x1))) b#(a(x1)) -> a#(c(b(b(x1)))) b#(b(x1)) -> a#(x1) TRS: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 TDG Processor: DPs: b#(a(x1)) -> b#(x1) b#(a(x1)) -> b#(b(x1)) b#(a(x1)) -> c#(b(b(x1))) b#(a(x1)) -> a#(c(b(b(x1)))) b#(b(x1)) -> a#(x1) TRS: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 graph: b#(a(x1)) -> b#(b(x1)) -> b#(b(x1)) -> a#(x1) b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> a#(c(b(b(x1)))) b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> c#(b(b(x1))) b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> b#(b(x1)) b#(a(x1)) -> b#(b(x1)) -> b#(a(x1)) -> b#(x1) b#(a(x1)) -> b#(x1) -> b#(b(x1)) -> a#(x1) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> a#(c(b(b(x1)))) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(b(b(x1))) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(b(x1)) b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 10/25 DPs: b#(a(x1)) -> b#(b(x1)) b#(a(x1)) -> b#(x1) TRS: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 interpretation: [b#](x0) = [3 3]x0 + [0], [-& 0 ] [0] [c](x0) = [0 0 ]x0 + [0], [1 0] [2] [b](x0) = [0 0]x0 + [0], [2 1] [3] [a](x0) = [0 0]x0 + [2] orientation: b#(a(x1)) = [5 4]x1 + [6] >= [4 3]x1 + [5] = b#(b(x1)) b#(a(x1)) = [5 4]x1 + [6] >= [3 3]x1 + [0] = b#(x1) [2 1] [3] a(x1) = [0 0]x1 + [2] >= x1 = x1 [3 2] [4] [3 2] [4] b(a(x1)) = [2 1]x1 + [3] >= [2 1]x1 + [3] = a(c(b(b(x1)))) [2 1] [3] [2 1] [3] b(b(x1)) = [1 0]x1 + [2] >= [0 0]x1 + [2] = a(x1) [0 0] [0] c(c(x1)) = [0 0]x1 + [0] >= x1 = x1 problem: DPs: TRS: a(x1) -> x1 b(a(x1)) -> a(c(b(b(x1)))) b(b(x1)) -> a(x1) c(c(x1)) -> x1 Qed