YES We show the termination of the TRS R: pred(s(x)) -> x minus(x,|0|()) -> x minus(x,s(y)) -> pred(minus(x,y)) quot(|0|(),s(y)) -> |0|() quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: minus#(x,s(y)) -> pred#(minus(x,y)) p2: minus#(x,s(y)) -> minus#(x,y) p3: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) p4: quot#(s(x),s(y)) -> minus#(x,y) and R consists of: r1: pred(s(x)) -> x r2: minus(x,|0|()) -> x r3: minus(x,s(y)) -> pred(minus(x,y)) r4: quot(|0|(),s(y)) -> |0|() r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) The estimated dependency graph contains the following SCCs: {p3} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: quot#(s(x),s(y)) -> quot#(minus(x,y),s(y)) and R consists of: r1: pred(s(x)) -> x r2: minus(x,|0|()) -> x r3: minus(x,s(y)) -> pred(minus(x,y)) r4: quot(|0|(),s(y)) -> |0|() r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) The set of usable rules consists of r1, r2, r3, r4 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: quot#_A(x1,x2) = max{1, x1 - 6} s_A(x1) = x1 + 16 minus_A(x1,x2) = x1 + 4 pred_A(x1) = max{3, x1 - 6} |0|_A = 7 quot_A(x1,x2) = max{2, x1 - 21, x2 - 8} base order 2: precedence: pred = quot > quot# = s > minus = |0| partial status: pi(quot#) = [] pi(s) = [1] pi(minus) = [] pi(pred) = [] pi(|0|) = [] pi(quot) = [] The following rules are removed: p1 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: minus#(x,s(y)) -> minus#(x,y) and R consists of: r1: pred(s(x)) -> x r2: minus(x,|0|()) -> x r3: minus(x,s(y)) -> pred(minus(x,y)) r4: quot(|0|(),s(y)) -> |0|() r5: quot(s(x),s(y)) -> s(quot(minus(x,y),s(y))) The set of usable rules consists of r4 Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: max/plus interpretation on natural numbers: minus#_A(x1,x2) = x2 s_A(x1) = x1 quot_A(x1,x2) = max{1, x1 - 1, x2 - 2} |0|_A = 0 base order 2: precedence: quot > s > minus# = |0| partial status: pi(minus#) = [2] pi(s) = [1] pi(quot) = [] pi(|0|) = [] The following rules are removed: p1 Then no dependency pair remains.