YES(O(1),O(n^1)) We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { implies(x, or(y, z)) -> or(y, implies(x, z)) , implies(not(x), y) -> or(x, y) , implies(not(x), or(y, z)) -> implies(y, or(x, z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { implies(not(x), y) -> or(x, y) , implies(not(x), or(y, z)) -> implies(y, or(x, z)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [implies](x1, x2) = [2] x1 + [2] x2 + [0] [not](x1) = [1] x1 + [2] [or](x1, x2) = [1] x1 + [1] x2 + [0] This order satisfies the following ordering constraints: [implies(x, or(y, z))] = [2] x + [2] y + [2] z + [0] >= [2] x + [1] y + [2] z + [0] = [or(y, implies(x, z))] [implies(not(x), y)] = [2] x + [2] y + [4] > [1] x + [1] y + [0] = [or(x, y)] [implies(not(x), or(y, z))] = [2] x + [2] y + [2] z + [4] > [2] x + [2] y + [2] z + [0] = [implies(y, or(x, z))] We return to the main proof. We are left with following problem, upon which TcT provides the certificate YES(O(1),O(n^1)). Strict Trs: { implies(x, or(y, z)) -> or(y, implies(x, z)) } Weak Trs: { implies(not(x), y) -> or(x, y) , implies(not(x), or(y, z)) -> implies(y, or(x, z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(n^1)) We use the processor 'matrix interpretation of dimension 1' to orient following rules strictly. Trs: { implies(x, or(y, z)) -> or(y, implies(x, z)) } The induced complexity on above rules (modulo remaining rules) is YES(?,O(n^1)) . These rules are moved into the corresponding weak component(s). Sub-proof: ---------- TcT has computed the following constructor-based matrix interpretation satisfying not(EDA). [implies](x1, x2) = [2] x1 + [2] x2 + [3] [not](x1) = [1] x1 + [1] [or](x1, x2) = [1] x1 + [1] x2 + [1] This order satisfies the following ordering constraints: [implies(x, or(y, z))] = [2] x + [2] y + [2] z + [5] > [2] x + [1] y + [2] z + [4] = [or(y, implies(x, z))] [implies(not(x), y)] = [2] x + [2] y + [5] > [1] x + [1] y + [1] = [or(x, y)] [implies(not(x), or(y, z))] = [2] x + [2] y + [2] z + [7] > [2] x + [2] y + [2] z + [5] = [implies(y, or(x, z))] 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: { implies(x, or(y, z)) -> or(y, implies(x, z)) , implies(not(x), y) -> or(x, y) , implies(not(x), or(y, z)) -> implies(y, or(x, z)) } Obligation: innermost runtime complexity Answer: YES(O(1),O(1)) Empty rules are trivially bounded Hurray, we answered YES(O(1),O(n^1))