YES(?,O(n^2)) Problem: a(a(x1)) -> d(c(x1)) a(b(x1)) -> c(c(c(x1))) b(b(x1)) -> a(c(c(x1))) c(c(x1)) -> b(x1) c(d(x1)) -> a(a(x1)) d(d(x1)) -> b(a(c(x1))) Proof: RT Transformation Processor: strict: a(a(x1)) -> d(c(x1)) a(b(x1)) -> c(c(c(x1))) b(b(x1)) -> a(c(c(x1))) c(c(x1)) -> b(x1) c(d(x1)) -> a(a(x1)) d(d(x1)) -> b(a(c(x1))) weak: Matrix Interpretation Processor: dimension: 1 interpretation: [b](x0) = x0 + 2, [d](x0) = x0 + 24, [c](x0) = x0, [a](x0) = x0 orientation: a(a(x1)) = x1 >= x1 + 24 = d(c(x1)) a(b(x1)) = x1 + 2 >= x1 = c(c(c(x1))) b(b(x1)) = x1 + 4 >= x1 = a(c(c(x1))) c(c(x1)) = x1 >= x1 + 2 = b(x1) c(d(x1)) = x1 + 24 >= x1 = a(a(x1)) d(d(x1)) = x1 + 48 >= x1 + 2 = b(a(c(x1))) problem: strict: a(a(x1)) -> d(c(x1)) c(c(x1)) -> b(x1) weak: a(b(x1)) -> c(c(c(x1))) b(b(x1)) -> a(c(c(x1))) c(d(x1)) -> a(a(x1)) d(d(x1)) -> b(a(c(x1))) Matrix Interpretation Processor: dimension: 1 interpretation: [b](x0) = x0 + 15, [d](x0) = x0 + 19, [c](x0) = x0 + 8, [a](x0) = x0 + 9 orientation: a(a(x1)) = x1 + 18 >= x1 + 27 = d(c(x1)) c(c(x1)) = x1 + 16 >= x1 + 15 = b(x1) a(b(x1)) = x1 + 24 >= x1 + 24 = c(c(c(x1))) b(b(x1)) = x1 + 30 >= x1 + 25 = a(c(c(x1))) c(d(x1)) = x1 + 27 >= x1 + 18 = a(a(x1)) d(d(x1)) = x1 + 38 >= x1 + 32 = b(a(c(x1))) problem: strict: a(a(x1)) -> d(c(x1)) weak: c(c(x1)) -> b(x1) a(b(x1)) -> c(c(c(x1))) b(b(x1)) -> a(c(c(x1))) c(d(x1)) -> a(a(x1)) d(d(x1)) -> b(a(c(x1))) Matrix Interpretation Processor: dimension: 2 interpretation: [1 6] [3] [b](x0) = [0 1]x0 + [2], [1 7] [4] [d](x0) = [0 1]x0 + [3], [1 3] [1] [c](x0) = [0 1]x0 + [1], [1 5] [2] [a](x0) = [0 1]x0 + [2] orientation: [1 10] [14] [1 10] [12] a(a(x1)) = [0 1 ]x1 + [4 ] >= [0 1 ]x1 + [4 ] = d(c(x1)) [1 6] [5] [1 6] [3] c(c(x1)) = [0 1]x1 + [2] >= [0 1]x1 + [2] = b(x1) [1 11] [15] [1 9] [12] a(b(x1)) = [0 1 ]x1 + [4 ] >= [0 1]x1 + [3 ] = c(c(c(x1))) [1 12] [18] [1 11] [17] b(b(x1)) = [0 1 ]x1 + [4 ] >= [0 1 ]x1 + [4 ] = a(c(c(x1))) [1 10] [14] [1 10] [14] c(d(x1)) = [0 1 ]x1 + [4 ] >= [0 1 ]x1 + [4 ] = a(a(x1)) [1 14] [29] [1 14] [29] d(d(x1)) = [0 1 ]x1 + [6 ] >= [0 1 ]x1 + [5 ] = b(a(c(x1))) problem: strict: weak: a(a(x1)) -> d(c(x1)) c(c(x1)) -> b(x1) a(b(x1)) -> c(c(c(x1))) b(b(x1)) -> a(c(c(x1))) c(d(x1)) -> a(a(x1)) d(d(x1)) -> b(a(c(x1))) Qed