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