YES Problem: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) Proof: DP Processor: DPs: a#(c(x1)) -> b#(c(c(x1))) b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) TDG Processor: DPs: a#(c(x1)) -> b#(c(c(x1))) b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) graph: b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> a#(b(x1)) b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) -> a#(c(x1)) -> b#(c(c(x1))) a#(c(x1)) -> b#(c(c(x1))) -> b#(c(x1)) -> a#(b(x1)) a#(c(x1)) -> b#(c(c(x1))) -> b#(c(x1)) -> b#(x1) Arctic Interpretation Processor: dimension: 1 usable rules: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) interpretation: [b#](x0) = -8x0 + 1, [a](x0) = 1x0 + 2, [a#](x0) = -6x0 + 1, [b](x0) = -1x0 + 1, [c](x0) = 1x0 + 8 orientation: a#(c(x1)) = -5x1 + 2 >= -6x1 + 1 = b#(c(c(x1))) b#(c(x1)) = -7x1 + 1 >= -8x1 + 1 = b#(x1) b#(c(x1)) = -7x1 + 1 >= -7x1 + 1 = a#(b(x1)) a(b(x1)) = x1 + 2 >= x1 = x1 a(c(x1)) = 2x1 + 9 >= 2x1 + 9 = c(b(c(c(x1)))) b(c(x1)) = x1 + 7 >= x1 + 2 = a(b(x1)) problem: DPs: b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) Restore Modifier: DPs: b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) EDG Processor: DPs: b#(c(x1)) -> b#(x1) b#(c(x1)) -> a#(b(x1)) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) graph: b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> b#(x1) b#(c(x1)) -> b#(x1) -> b#(c(x1)) -> a#(b(x1)) SCC Processor: #sccs: 1 #rules: 1 #arcs: 2/4 DPs: b#(c(x1)) -> b#(x1) TRS: a(b(x1)) -> x1 a(c(x1)) -> c(b(c(c(x1)))) b(c(x1)) -> a(b(x1)) Usable Rule Processor: DPs: b#(c(x1)) -> b#(x1) TRS: Arctic Interpretation Processor: dimension: 4 usable rules: interpretation: [b#](x0) = [0 -& -& -&]x0, [1 0 0 0] [0] [0 0 0 0] [0] [c](x0) = [1 0 0 0]x0 + [0] [0 0 0 0] [0] orientation: b#(c(x1)) = [1 0 0 0]x1 + [0] >= [0 -& -& -&]x1 = b#(x1) problem: DPs: TRS: Qed