YES We show the termination of the TRS R: not(and(x,y)) -> or(not(x),not(y)) not(or(x,y)) -> and(not(x),not(y)) and(x,or(y,z)) -> or(and(x,y),and(x,z)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: not#(and(x,y)) -> not#(x) p2: not#(and(x,y)) -> not#(y) p3: not#(or(x,y)) -> and#(not(x),not(y)) p4: not#(or(x,y)) -> not#(x) p5: not#(or(x,y)) -> not#(y) p6: and#(x,or(y,z)) -> and#(x,y) p7: and#(x,or(y,z)) -> and#(x,z) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,z)) The estimated dependency graph contains the following SCCs: {p1, p2, p4, p5} {p6, p7} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: not#(and(x,y)) -> not#(x) p2: not#(or(x,y)) -> not#(y) p3: not#(or(x,y)) -> not#(x) p4: not#(and(x,y)) -> not#(y) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: not#_A(x1) = x1 + 3 and_A(x1,x2) = max{x1, x2} or_A(x1,x2) = max{x1, x2 + 1} precedence: not# = and = or partial status: pi(not#) = [] pi(and) = [2] pi(or) = [2] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: not#(and(x,y)) -> not#(x) p2: not#(or(x,y)) -> not#(x) p3: not#(and(x,y)) -> not#(y) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,z)) 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: not#(and(x,y)) -> not#(x) p2: not#(and(x,y)) -> not#(y) p3: not#(or(x,y)) -> not#(x) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: not#_A(x1) = max{3, x1 + 2} and_A(x1,x2) = max{x1, x2 + 2} or_A(x1,x2) = max{x1 + 1, x2 + 1} precedence: not# = and = or partial status: pi(not#) = [] pi(and) = [2] pi(or) = [2] The next rules are strictly ordered: p2 We remove them from the problem. -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: not#(and(x,y)) -> not#(x) p2: not#(or(x,y)) -> not#(x) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(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: not#(and(x,y)) -> not#(x) p2: not#(or(x,y)) -> not#(x) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: not#_A(x1) = x1 + 2 and_A(x1,x2) = max{x1 + 2, x2 + 2} or_A(x1,x2) = max{x1, x2} precedence: not# = and = or partial status: pi(not#) = [] pi(and) = [2] pi(or) = [2] 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: not#(or(x,y)) -> not#(x) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(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: not#(or(x,y)) -> not#(x) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: not#_A(x1) = x1 + 3 or_A(x1,x2) = max{x1 + 1, x2} precedence: not# = or partial status: pi(not#) = [] pi(or) = [2] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: and#(x,or(y,z)) -> and#(x,y) p2: and#(x,or(y,z)) -> and#(x,z) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: and#_A(x1,x2) = max{0, x2 - 2} or_A(x1,x2) = max{3, x1 + 1, x2} precedence: and# = or partial status: pi(and#) = [] pi(or) = [2] 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: and#(x,or(y,z)) -> and#(x,z) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(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: and#(x,or(y,z)) -> and#(x,z) and R consists of: r1: not(and(x,y)) -> or(not(x),not(y)) r2: not(or(x,y)) -> and(not(x),not(y)) r3: and(x,or(y,z)) -> or(and(x,y),and(x,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: and#_A(x1,x2) = max{0, x2 - 2} or_A(x1,x2) = max{x1 + 3, x2 + 1} precedence: and# = or partial status: pi(and#) = [] pi(or) = [2] The next rules are strictly ordered: p1 We remove them from the problem. Then no dependency pair remains.