We consider the system 469. Alphabet: all : [term -> form] --> form and : [form * form] --> form or : [form * form] --> form Rules: all(/\x.X) => X all(/\x.and(X(x), Y(x))) => and(all(X), all(Y)) all(/\x.or(X(x), Y)) => or(all(X), Y) all(/\x.or(X, Y(x))) => or(X, all(Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): all(/\x.X) >? X all(/\x.and(X(x), Y(x))) >? and(all(X), all(Y)) all(/\x.or(X(x), Y)) >? or(all(X), Y) all(/\x.or(X, Y(x))) >? or(X, all(Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: all = \G0.3G0(0) and = \y0y1.3 + y0 + y1 or = \y0y1.3 + y0 + y1 Using this interpretation, the requirements translate to: [[all(/\x._x0)]] = 3x0 >= x0 = [[_x0]] [[all(/\x.and(_x0(x), _x1(x)))]] = 9 + 3F0(0) + 3F1(0) > 3 + 3F0(0) + 3F1(0) = [[and(all(_x0), all(_x1))]] [[all(/\x.or(_x0(x), _x1))]] = 9 + 3x1 + 3F0(0) > 3 + x1 + 3F0(0) = [[or(all(_x0), _x1)]] [[all(/\x.or(_x0, _x1(x)))]] = 9 + 3x0 + 3F1(0) > 3 + x0 + 3F1(0) = [[or(_x0, all(_x1))]] We can thus remove the following rules: all(/\x.and(X(x), Y(x))) => and(all(X), all(Y)) all(/\x.or(X(x), Y)) => or(all(X), Y) all(/\x.or(X, Y(x))) => or(X, all(Y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): all(/\x.X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: all = \G0.1 + G0(0) Using this interpretation, the requirements translate to: [[all(/\x._x0)]] = 1 + x0 > x0 = [[_x0]] We can thus remove the following rules: all(/\x.X) => X All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.