/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: f(a(),f(a(),x)) -> f(a(),f(x,f(f(a(),a()),a()))) Proof: Extended Uncurrying Processor: application symbol: f symbol table: a ==> a0/0 a1/1 a2/2 uncurry-rules: f(a1(x2),x3) -> a2(x2,x3) f(a0(),x2) -> a1(x2) eta-rules: f(f(a(),f(a(),x)),x1) -> f(f(a(),f(x,f(f(a(),a()),a()))),x1) problem: a1(a1(x)) -> a1(f(x,a2(a0(),a0()))) a2(a1(x),x1) -> a2(f(x,a2(a0(),a0())),x1) f(a1(x2),x3) -> a2(x2,x3) f(a0(),x2) -> a1(x2) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1 0 0] [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [1 0 0] , [1 1 0] [0] [a1](x0) = [0 0 0]x0 + [1] [1 0 0] [0], [0] [a0] = [1] [0], [1 0 0] [1 1 1] [f](x0, x1) = [0 1 0]x0 + [0 0 1]x1 [0 1 0] [1 0 1] orientation: [1 1 0] [1] [1 1 0] [0] a1(a1(x)) = [0 0 0]x + [1] >= [0 0 0]x + [1] = a1(f(x,a2(a0(),a0()))) [1 1 0] [0] [1 0 0] [0] [1 1 0] [1 0 0] [1 0 0] [1 0 0] a2(a1(x),x1) = [0 0 0]x + [0 0 0]x1 >= [0 0 0]x + [0 0 0]x1 = a2(f(x,a2(a0(),a0())),x1) [0 0 0] [1 0 0] [0 0 0] [1 0 0] [1 1 0] [1 1 1] [0] [1 0 0] [1 0 0] f(a1(x2),x3) = [0 0 0]x2 + [0 0 1]x3 + [1] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3) [0 0 0] [1 0 1] [1] [0 0 0] [1 0 0] [1 1 1] [0] [1 1 0] [0] f(a0(),x2) = [0 0 1]x2 + [1] >= [0 0 0]x2 + [1] = a1(x2) [1 0 1] [1] [1 0 0] [0] problem: a2(a1(x),x1) -> a2(f(x,a2(a0(),a0())),x1) f(a1(x2),x3) -> a2(x2,x3) f(a0(),x2) -> a1(x2) Matrix Interpretation Processor: dim=3 interpretation: [1 1 0] [1 0 0] [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [0 0 0] , [1 0 1] [0] [a1](x0) = [0 1 0]x0 + [1] [0 1 0] [0], [0] [a0] = [1] [1], [1 0 1] [1 0 1] [0] [f](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [0] [1 0 0] [0 1 1] [1] orientation: [1 1 1] [1 0 0] [1] [1 1 1] [1 0 0] [1] a2(a1(x),x1) = [0 0 0]x + [0 0 0]x1 + [0] >= [0 0 0]x + [0 0 0]x1 + [0] = a2(f(x,a2(a0(),a0())),x1) [0 0 0] [0 0 0] [0] [0 0 0] [0 0 0] [0] [1 1 1] [1 0 1] [0] [1 1 0] [1 0 0] f(a1(x2),x3) = [0 1 0]x2 + [0 1 0]x3 + [1] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3) [1 0 1] [0 1 1] [1] [0 0 0] [0 0 0] [1 0 1] [1] [1 0 1] [0] f(a0(),x2) = [0 1 0]x2 + [1] >= [0 1 0]x2 + [1] = a1(x2) [0 1 1] [1] [0 1 0] [0] problem: a2(a1(x),x1) -> a2(f(x,a2(a0(),a0())),x1) f(a1(x2),x3) -> a2(x2,x3) Matrix Interpretation Processor: dim=3 interpretation: [1 0 0] [1 0 0] [a2](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [0 0 0] , [1 0 0] [1] [a1](x0) = [0 0 0]x0 + [0] [0 0 0] [0], [0] [a0] = [0] [0], [1 0 0] [1 0 0] [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1 [0 0 0] [0 0 0] orientation: [1 0 0] [1 0 0] [1] [1 0 0] [1 0 0] a2(a1(x),x1) = [0 0 0]x + [0 0 0]x1 + [0] >= [0 0 0]x + [0 0 0]x1 = a2(f(x,a2(a0(),a0())),x1) [0 0 0] [0 0 0] [0] [0 0 0] [0 0 0] [1 0 0] [1 0 0] [1] [1 0 0] [1 0 0] f(a1(x2),x3) = [0 0 0]x2 + [0 0 0]x3 + [0] >= [0 0 0]x2 + [0 0 0]x3 = a2(x2,x3) [0 0 0] [0 0 0] [0] [0 0 0] [0 0 0] problem: Qed