YES We show the termination of the TRS R: *(x,+(y,z)) -> +(*(x,y),*(x,z)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: *#(x,+(y,z)) -> *#(x,y) p2: *#(x,+(y,z)) -> *#(x,z) and R consists of: r1: *(x,+(y,z)) -> +(*(x,y),*(x,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: *#(x,+(y,z)) -> *#(x,y) p2: *#(x,+(y,z)) -> *#(x,z) and R consists of: r1: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *#_A(x1,x2) = x1 + x2 + 2 +_A(x1,x2) = x1 + x2 + 1 base order 2: precedence: *# > + partial status: pi(*#) = [1] pi(+) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: *#(x,+(y,z)) -> *#(x,z) and R consists of: r1: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: *#(x,+(y,z)) -> *#(x,z) and R consists of: r1: *(x,+(y,z)) -> +(*(x,y),*(x,z)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >MG) base order 1: matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: *#_A(x1,x2) = x2 +_A(x1,x2) = x2 + 1 base order 2: precedence: + > *# partial status: pi(*#) = [] pi(+) = [2] The following rules are removed: p1 Then no dependency pair remains.