YES(O(1),O(n^2)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(0()) -> true() , f(1()) -> false() , f(s(x)) -> f(x) , if(true(), x, y) -> x , if(false(), x, y) -> y , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { if(false(), x, y) -> y } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [0 0] [0] [0] = [0] [0] [true] = [0] [0] [1] = [2] [0] [false] = [2] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [if](x1, x2, x3) = [1 0] x1 + [2 0] x2 + [1 0] x3 + [0] [0 0] [0 2] [0 2] [0] [g](x1, x2) = [3 0] x1 + [1 3] x2 + [0] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [0] This order satisfies the following ordering constraints: [f(0())] = [0] [0] >= [0] [0] = [true()] [f(1())] = [2] [0] >= [2] [0] = [false()] [f(s(x))] = [1 0] x + [0] [0 0] [0] >= [1 0] x + [0] [0 0] [0] = [f(x)] [if(true(), x, y)] = [2 0] x + [1 0] y + [0] [0 2] [0 2] [0] >= [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [2 0] x + [1 0] y + [2] [0 2] [0 2] [0] > [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [3 0] x + [4 3] y + [0] [0 0] [0 0] [0] >= [3 0] x + [4 3] y + [0] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [3 0] x + [1 0] y + [0] [0 0] [0 0] [0] >= [3 0] x + [1 0] y + [0] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(0()) -> true() , f(1()) -> false() , f(s(x)) -> f(x) , if(true(), x, y) -> x , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Weak Trs: { if(false(), x, y) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { if(true(), x, y) -> x } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [1 0] [0] [0] = [2] [0] [true] = [2] [2] [1] = [0] [0] [false] = [0] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [if](x1, x2, x3) = [1 1] x1 + [1 0] x2 + [1 0] x3 + [0] [0 0] [0 2] [0 2] [0] [g](x1, x2) = [3 0] x1 + [1 3] x2 + [0] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [0] This order satisfies the following ordering constraints: [f(0())] = [2] [2] >= [2] [2] = [true()] [f(1())] = [0] [0] >= [0] [0] = [false()] [f(s(x))] = [1 0] x + [0] [1 0] [0] >= [1 0] x + [0] [1 0] [0] = [f(x)] [if(true(), x, y)] = [1 0] x + [1 0] y + [4] [0 2] [0 2] [0] > [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [1 0] x + [1 0] y + [0] [0 2] [0 2] [0] >= [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [3 0] x + [4 3] y + [0] [0 0] [0 0] [0] >= [3 0] x + [4 3] y + [0] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [3 0] x + [1 0] y + [0] [0 0] [0 0] [0] >= [3 0] x + [1 0] y + [0] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(0()) -> true() , f(1()) -> false() , f(s(x)) -> f(x) , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Weak Trs: { if(true(), x, y) -> x , if(false(), x, y) -> y } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [0 0] [0] [0] = [0] [0] [true] = [0] [0] [1] = [0] [0] [false] = [0] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [if](x1, x2, x3) = [1 0] x1 + [2 0] x2 + [1 0] x3 + [0] [0 0] [0 2] [0 1] [0] [g](x1, x2) = [3 0] x1 + [1 3] x2 + [2] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [1] This order satisfies the following ordering constraints: [f(0())] = [0] [0] >= [0] [0] = [true()] [f(1())] = [0] [0] >= [0] [0] = [false()] [f(s(x))] = [1 0] x + [0] [0 0] [0] >= [1 0] x + [0] [0 0] [0] = [f(x)] [if(true(), x, y)] = [2 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [2 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [3 0] x + [4 3] y + [5] [0 0] [0 0] [0] > [3 0] x + [4 3] y + [4] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [3 0] x + [1 0] y + [2] [0 0] [0 0] [0] > [3 0] x + [1 0] y + [0] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(0()) -> true() , f(1()) -> false() , f(s(x)) -> f(x) } Weak Trs: { if(true(), x, y) -> x , if(false(), x, y) -> y , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { f(0()) -> true() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [0 0] [0] [0] = [1] [0] [true] = [0] [0] [1] = [0] [0] [false] = [0] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [if](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0] [0 0] [0 2] [0 1] [0] [g](x1, x2) = [2 0] x1 + [1 2] x2 + [0] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [2] This order satisfies the following ordering constraints: [f(0())] = [1] [0] > [0] [0] = [true()] [f(1())] = [0] [0] >= [0] [0] = [false()] [f(s(x))] = [1 0] x + [0] [0 0] [0] >= [1 0] x + [0] [0 0] [0] = [f(x)] [if(true(), x, y)] = [1 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [1 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [2 0] x + [3 2] y + [4] [0 0] [0 0] [0] > [2 0] x + [3 2] y + [0] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [2 0] x + [1 0] y + [0] [0 0] [0 0] [0] >= [2 0] x + [1 0] y + [0] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(1()) -> false() , f(s(x)) -> f(x) } Weak Trs: { f(0()) -> true() , if(true(), x, y) -> x , if(false(), x, y) -> y , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { f(1()) -> false() } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [0 0] [0] [0] = [0] [0] [true] = [0] [0] [1] = [1] [0] [false] = [0] [0] [s](x1) = [1 0] x1 + [0] [0 0] [0] [if](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0] [0 0] [0 2] [0 1] [0] [g](x1, x2) = [2 0] x1 + [1 2] x2 + [0] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [0] This order satisfies the following ordering constraints: [f(0())] = [0] [0] >= [0] [0] = [true()] [f(1())] = [1] [0] > [0] [0] = [false()] [f(s(x))] = [1 0] x + [0] [0 0] [0] >= [1 0] x + [0] [0 0] [0] = [f(x)] [if(true(), x, y)] = [1 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [1 0] x + [1 0] y + [0] [0 2] [0 1] [0] >= [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [2 0] x + [3 2] y + [0] [0 0] [0 0] [0] >= [2 0] x + [3 2] y + [0] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [2 0] x + [1 0] y + [0] [0 0] [0 0] [0] >= [2 0] x + [1 0] y + [0] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^2)). Strict Trs: { f(s(x)) -> f(x) } Weak Trs: { f(0()) -> true() , f(1()) -> false() , if(true(), x, y) -> x , if(false(), x, y) -> y , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^2)) We use the processor 'matrix interpretation of dimension 2' to orient following rules strictly. Trs: { f(s(x)) -> f(x) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^2)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [f](x1) = [1 0] x1 + [0] [0 0] [0] [0] = [0] [0] [true] = [0] [0] [1] = [0] [0] [false] = [0] [0] [s](x1) = [1 0] x1 + [1] [0 0] [0] [if](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 0] x3 + [0] [0 0] [0 1] [0 1] [0] [g](x1, x2) = [2 0] x1 + [1 2] x2 + [0] [0 0] [0 0] [0] [c](x1) = [1 0] x1 + [0] [1 1] [2] This order satisfies the following ordering constraints: [f(0())] = [0] [0] >= [0] [0] = [true()] [f(1())] = [0] [0] >= [0] [0] = [false()] [f(s(x))] = [1 0] x + [1] [0 0] [0] > [1 0] x + [0] [0 0] [0] = [f(x)] [if(true(), x, y)] = [1 0] x + [1 0] y + [0] [0 1] [0 1] [0] >= [1 0] x + [0] [0 1] [0] = [x] [if(false(), x, y)] = [1 0] x + [1 0] y + [0] [0 1] [0 1] [0] >= [1 0] y + [0] [0 1] [0] = [y] [g(x, c(y))] = [2 0] x + [3 2] y + [4] [0 0] [0 0] [0] > [2 0] x + [3 2] y + [2] [0 0] [0 0] [0] = [g(x, g(s(c(y)), y))] [g(s(x), s(y))] = [2 0] x + [1 0] y + [3] [0 0] [0 0] [0] > [2 0] x + [1 0] y + [2] [0 0] [0 0] [0] = [if(f(x), s(x), s(y))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { f(0()) -> true() , f(1()) -> false() , f(s(x)) -> f(x) , if(true(), x, y) -> x , if(false(), x, y) -> y , g(x, c(y)) -> g(x, g(s(c(y)), y)) , g(s(x), s(y)) -> if(f(x), s(x), s(y)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^2))