/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(a(b(x1))) -> b(b(a(a(x1)))) b(x1) -> c(a(x1)) Proof: String Reversal Processor: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) DP Processor: DPs: b#(a(a(x1))) -> b#(x1) b#(a(a(x1))) -> b#(b(x1)) b#(a(a(x1))) -> a#(b(b(x1))) b#(a(a(x1))) -> a#(a(b(b(x1)))) b#(x1) -> a#(c(x1)) TRS: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) TDG Processor: DPs: b#(a(a(x1))) -> b#(x1) b#(a(a(x1))) -> b#(b(x1)) b#(a(a(x1))) -> a#(b(b(x1))) b#(a(a(x1))) -> a#(a(b(b(x1)))) b#(x1) -> a#(c(x1)) TRS: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) graph: b#(a(a(x1))) -> b#(b(x1)) -> b#(x1) -> a#(c(x1)) b#(a(a(x1))) -> b#(b(x1)) -> b#(a(a(x1))) -> a#(a(b(b(x1)))) b#(a(a(x1))) -> b#(b(x1)) -> b#(a(a(x1))) -> a#(b(b(x1))) b#(a(a(x1))) -> b#(b(x1)) -> b#(a(a(x1))) -> b#(b(x1)) b#(a(a(x1))) -> b#(b(x1)) -> b#(a(a(x1))) -> b#(x1) b#(a(a(x1))) -> b#(x1) -> b#(x1) -> a#(c(x1)) b#(a(a(x1))) -> b#(x1) -> b#(a(a(x1))) -> a#(a(b(b(x1)))) b#(a(a(x1))) -> b#(x1) -> b#(a(a(x1))) -> a#(b(b(x1))) b#(a(a(x1))) -> b#(x1) -> b#(a(a(x1))) -> b#(b(x1)) b#(a(a(x1))) -> b#(x1) -> b#(a(a(x1))) -> b#(x1) SCC Processor: #sccs: 1 #rules: 2 #arcs: 10/25 DPs: b#(a(a(x1))) -> b#(b(x1)) b#(a(a(x1))) -> b#(x1) TRS: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) Arctic Interpretation Processor: dimension: 2 usable rules: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) interpretation: [b#](x0) = [0 0]x0 + [0], [0 -&] [0 ] [c](x0) = [-& -&]x0 + [-&], [0 0] [0] [b](x0) = [0 0]x0 + [0], [0 1] [0] [a](x0) = [0 0]x0 + [0] orientation: b#(a(a(x1))) = [1 1]x1 + [1] >= [0 0]x1 + [0] = b#(b(x1)) b#(a(a(x1))) = [1 1]x1 + [1] >= [0 0]x1 + [0] = b#(x1) [0 1] [0] a(x1) = [0 0]x1 + [0] >= x1 = x1 [1 1] [1] [1 1] [1] b(a(a(x1))) = [1 1]x1 + [1] >= [1 1]x1 + [1] = a(a(b(b(x1)))) [0 0] [0] [0 -&] [0] b(x1) = [0 0]x1 + [0] >= [0 -&]x1 + [0] = a(c(x1)) problem: DPs: TRS: a(x1) -> x1 b(a(a(x1))) -> a(a(b(b(x1)))) b(x1) -> a(c(x1)) Qed