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: lexicographic combination of reduction pairs: 1. matrix interpretations: carrier: N^2 order: standard order interpretations: quot#_A(x1,x2,x3) = ((1,1),(1,0)) x1 s_A(x1) = ((1,1),(1,1)) x1 + (1,1) |0|_A() = (1,1) 2. matrix interpretations: carrier: N^2 order: standard order interpretations: quot#_A(x1,x2,x3) = ((1,1),(1,1)) x1 s_A(x1) = ((1,1),(1,1)) x1 + (1,1) |0|_A() = (1,1) 3. matrix interpretations: carrier: N^2 order: standard order interpretations: quot#_A(x1,x2,x3) = ((1,1),(0,0)) x1 s_A(x1) = ((1,1),(1,1)) x1 + (1,1) |0|_A() = (1,1) 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)