We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We add the following weak dependency pairs: Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } and mark the set of starting terms. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) , quot(0(), s(y)) -> 0() , quot(s(x), s(y)) -> s(quot(minus(x, y), s(y))) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We replace rewrite rules by usable rules: Strict Usable Rules: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Strict Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The weightgap principle applies (using the following constant growth matrix-interpretation) The following argument positions are usable: Uargs(c_2) = {1}, Uargs(quot^#) = {1}, Uargs(c_4) = {1} TcT has computed the following constructor-restricted matrix interpretation. [minus](x1, x2) = [1 1] x1 + [2] [2 2] [0] [0] = [0] [0] [s](x1) = [1 1] x1 + [0] [0 0] [2] [minus^#](x1, x2) = [0] [0] [c_1] = [0] [0] [c_2](x1) = [1 0] x1 + [0] [0 1] [0] [quot^#](x1, x2) = [1 0] x1 + [0] [0 0] [0] [c_3] = [0] [0] [c_4](x1) = [1 0] x1 + [0] [0 1] [0] The order satisfies the following ordering constraints: [minus(x, 0())] = [1 1] x + [2] [2 2] [0] > [1 0] x + [0] [0 1] [0] = [x] [minus(s(x), s(y))] = [1 1] x + [4] [2 2] [4] > [1 1] x + [2] [2 2] [0] = [minus(x, y)] [minus^#(x, 0())] = [0] [0] >= [0] [0] = [c_1()] [minus^#(s(x), s(y))] = [0] [0] >= [0] [0] = [c_2(minus^#(x, y))] [quot^#(0(), s(y))] = [0] [0] >= [0] [0] = [c_3()] [quot^#(s(x), s(y))] = [1 1] x + [0] [0 0] [0] ? [1 1] x + [2] [0 0] [0] = [c_4(quot^#(minus(x, y), s(y)))] Further, it can be verified that all rules not oriented are covered by the weightgap condition. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(x, 0()) -> c_1() , minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(0(), s(y)) -> c_3() , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We estimate the number of application of {1,3} by applications of Pre({1,3}) = {2,4}. Here rules are labeled as follows: DPs: { 1: minus^#(x, 0()) -> c_1() , 2: minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , 3: quot^#(0(), s(y)) -> c_3() , 4: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak DPs: { minus^#(x, 0()) -> c_1() , quot^#(0(), s(y)) -> c_3() } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { minus^#(x, 0()) -> c_1() , quot^#(0(), s(y)) -> c_3() } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'Small Polynomial Path Order (PS,1-bounded)' to orient following rules strictly. DPs: { 1: minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , 2: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Trs: { minus(s(x), s(y)) -> minus(x, y) } Sub-proof: ---------- The input was oriented with the instance of 'Small Polynomial Path Order (PS,1-bounded)' as induced by the safe mapping safe(minus) = {2}, safe(0) = {}, safe(s) = {1}, safe(minus^#) = {2}, safe(c_2) = {}, safe(quot^#) = {}, safe(c_4) = {} and precedence quot^# > minus . Following symbols are considered recursive: {minus^#, quot^#} The recursion depth is 1. Further, following argument filtering is employed: pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(minus^#) = [1], pi(c_2) = [1], pi(quot^#) = [1, 2], pi(c_4) = [1] Usable defined function symbols are a subset of: {minus, minus^#, quot^#} For your convenience, here are the satisfied ordering constraints: pi(minus^#(s(x), s(y))) = minus^#(s(; x);) > c_2(minus^#(x;);) = pi(c_2(minus^#(x, y))) pi(quot^#(s(x), s(y))) = quot^#(s(; x), s(; y);) > c_4(quot^#(x, s(; y););) = pi(c_4(quot^#(minus(x, y), s(y)))) pi(minus(x, 0())) = x >= x = pi(x) pi(minus(s(x), s(y))) = s(; x) > x = pi(minus(x, y)) The strictly oriented rules are moved into the weak component. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak DPs: { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) The following weak DPs constitute a sub-graph of the DG that is closed under successors. The DPs are removed. { minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) , quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) } We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Weak Trs: { minus(x, 0()) -> x , minus(s(x), s(y)) -> minus(x, y) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) No rule is usable, rules are removed from the input problem. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(1)). Rules: Empty Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))