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