/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: strict: b(c(a(x1))) -> d(d(x1)) b(x1) -> c(c(x1)) a(a(x1)) -> a(c(b(a(x1)))) weak: a(b(x1)) -> d(x1) d(x1) -> a(b(x1)) Proof: String Reversal Processor: strict: a(c(b(x1))) -> d(d(x1)) b(x1) -> c(c(x1)) a(a(x1)) -> a(b(c(a(x1)))) weak: b(a(x1)) -> d(x1) d(x1) -> b(a(x1)) Matrix Interpretation Processor: dim=3 interpretation: [1 0 1] [0] [d](x0) = [0 0 0]x0 + [2] [0 0 0] [0], [1 0 0] [0] [b](x0) = [0 0 1]x0 + [2] [0 0 0] [0], [1 0 0] [c](x0) = [0 0 0]x0 [0 2 0] , [1 0 1] [0] [a](x0) = [0 0 0]x0 + [2] [0 0 0] [0] orientation: [1 0 2] [4] [1 0 1] [0] a(c(b(x1))) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [2] = d(d(x1)) [0 0 0] [0] [0 0 0] [0] [1 0 0] [0] [1 0 0] b(x1) = [0 0 1]x1 + [2] >= [0 0 0]x1 = c(c(x1)) [0 0 0] [0] [0 0 0] [1 0 1] [0] [1 0 1] [0] a(a(x1)) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [2] = a(b(c(a(x1)))) [0 0 0] [0] [0 0 0] [0] [1 0 1] [0] [1 0 1] [0] b(a(x1)) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [2] = d(x1) [0 0 0] [0] [0 0 0] [0] [1 0 1] [0] [1 0 1] [0] d(x1) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [2] = b(a(x1)) [0 0 0] [0] [0 0 0] [0] problem: strict: b(x1) -> c(c(x1)) a(a(x1)) -> a(b(c(a(x1)))) weak: b(a(x1)) -> d(x1) d(x1) -> b(a(x1)) Arctic Interpretation Processor: dimension: 2 interpretation: [6 7] [d](x0) = [6 7]x0, [3 -&] [b](x0) = [2 0 ]x0, [1 -&] [c](x0) = [-& -&]x0, [3 4] [a](x0) = [6 7]x0 orientation: [3 -&] [2 -&] b(x1) = [2 0 ]x1 >= [-& -&]x1 = c(c(x1)) [10 11] [10 11] a(a(x1)) = [13 14]x1 >= [13 14]x1 = a(b(c(a(x1)))) [6 7] [6 7] b(a(x1)) = [6 7]x1 >= [6 7]x1 = d(x1) [6 7] [6 7] d(x1) = [6 7]x1 >= [6 7]x1 = b(a(x1)) problem: strict: a(a(x1)) -> a(b(c(a(x1)))) weak: b(a(x1)) -> d(x1) d(x1) -> b(a(x1)) Arctic Interpretation Processor: dimension: 2 interpretation: [0 3] [d](x0) = [1 7]x0, [b](x0) = x0, [0 1 ] [c](x0) = [0 -&]x0, [0 3] [a](x0) = [1 7]x0 orientation: [4 10] [3 8 ] a(a(x1)) = [8 14]x1 >= [7 10]x1 = a(b(c(a(x1)))) [0 3] [0 3] b(a(x1)) = [1 7]x1 >= [1 7]x1 = d(x1) [0 3] [0 3] d(x1) = [1 7]x1 >= [1 7]x1 = b(a(x1)) problem: strict: weak: b(a(x1)) -> d(x1) d(x1) -> b(a(x1)) Qed