YES We show the termination of the TRS R: quot(|0|(),s(y),s(z)) -> |0|() quot(s(x),s(y),z) -> quot(x,y,z) quot(x,|0|(),s(z)) -> s(quot(x,s(z),s(z))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: quot#(s(x),s(y),z) -> quot#(x,y,z) p2: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z)) and R consists of: r1: quot(|0|(),s(y),s(z)) -> |0|() r2: quot(s(x),s(y),z) -> quot(x,y,z) r3: quot(x,|0|(),s(z)) -> s(quot(x,s(z),s(z))) The estimated dependency graph contains the following SCCs: {p1, p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: quot#(s(x),s(y),z) -> quot#(x,y,z) p2: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z)) and R consists of: r1: quot(|0|(),s(y),s(z)) -> |0|() r2: quot(s(x),s(y),z) -> quot(x,y,z) r3: quot(x,|0|(),s(z)) -> s(quot(x,s(z),s(z))) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: max/plus interpretations on natural numbers: quot#_A(x1,x2,x3) = max{x1 + 1, x2 - 1, x3 + 1} s_A(x1) = max{3, x1} |0|_A = 2 precedence: quot# = s = |0| partial status: pi(quot#) = [1, 3] pi(s) = [1] pi(|0|) = [] The next rules are strictly ordered: p1 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: quot#(x,|0|(),s(z)) -> quot#(x,s(z),s(z)) and R consists of: r1: quot(|0|(),s(y),s(z)) -> |0|() r2: quot(s(x),s(y),z) -> quot(x,y,z) r3: quot(x,|0|(),s(z)) -> s(quot(x,s(z),s(z))) The estimated dependency graph contains the following SCCs: (no SCCs)