YES We show the termination of the TRS R: +(*(x,y),*(x,z)) -> *(x,+(y,z)) +(+(x,y),z) -> +(x,+(y,z)) +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: +#(*(x,y),*(x,z)) -> +#(y,z) p2: +#(+(x,y),z) -> +#(x,+(y,z)) p3: +#(+(x,y),z) -> +#(y,z) p4: +#(*(x,y),+(*(x,z),u())) -> +#(*(x,+(y,z)),u()) p5: +#(*(x,y),+(*(x,z),u())) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The estimated dependency graph contains the following SCCs: {p1, p2, p3, p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: +#(*(x,y),*(x,z)) -> +#(y,z) p2: +#(*(x,y),+(*(x,z),u())) -> +#(y,z) p3: +#(+(x,y),z) -> +#(y,z) p4: +#(+(x,y),z) -> +#(x,+(y,z)) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The set of usable rules consists of r1, r2, r3 Take the reduction pair: weighted path order base order: max/plus interpretations on natural numbers: +#_A(x1,x2) = x1 + 1 *_A(x1,x2) = max{x1 + 4, x2 + 1} +_A(x1,x2) = max{x1, x2} u_A = 1 precedence: + = u > +# > * partial status: pi(+#) = [] pi(*) = [2] pi(+) = [] pi(u) = [] 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: +#(*(x,y),+(*(x,z),u())) -> +#(y,z) p2: +#(+(x,y),z) -> +#(y,z) p3: +#(+(x,y),z) -> +#(x,+(y,z)) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The estimated dependency graph contains the following SCCs: {p1, p2, p3} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: +#(*(x,y),+(*(x,z),u())) -> +#(y,z) p2: +#(+(x,y),z) -> +#(x,+(y,z)) p3: +#(+(x,y),z) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The set of usable rules consists of r1, r2, r3 Take the reduction pair: weighted path order base order: max/plus interpretations on natural numbers: +#_A(x1,x2) = x1 + 4 *_A(x1,x2) = max{x1 + 8, x2 + 5} +_A(x1,x2) = max{3, x1, x2} u_A = 4 precedence: +# = * = + = u partial status: pi(+#) = [] pi(*) = [] pi(+) = [] pi(u) = [] 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: +#(+(x,y),z) -> +#(x,+(y,z)) p2: +#(+(x,y),z) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) 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,z)) p2: +#(+(x,y),z) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The set of usable rules consists of r1, r2, r3 Take the reduction pair: weighted path order base order: max/plus interpretations on natural numbers: +#_A(x1,x2) = x1 + 5 +_A(x1,x2) = max{x1 + 2, x2} *_A(x1,x2) = x1 + 3 u_A = 2 precedence: +# = + = * = u partial status: pi(+#) = [] pi(+) = [] pi(*) = [1] pi(u) = [] 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: +#(+(x,y),z) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) 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) -> +#(y,z) and R consists of: r1: +(*(x,y),*(x,z)) -> *(x,+(y,z)) r2: +(+(x,y),z) -> +(x,+(y,z)) r3: +(*(x,y),+(*(x,z),u())) -> +(*(x,+(y,z)),u()) The set of usable rules consists of (no rules) Take the reduction pair: weighted path order base order: max/plus interpretations on natural numbers: +#_A(x1,x2) = max{x1 + 1, x2 + 1} +_A(x1,x2) = x2 precedence: +# = + partial status: pi(+#) = [1] pi(+) = [2] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.