YES We show the termination of the TRS R: app(app(and(),true()),true()) -> true() app(app(and(),x),false()) -> false() app(app(and(),false()),y) -> false() app(app(or(),true()),y) -> true() app(app(or(),x),true()) -> true() app(app(or(),false()),false()) -> false() app(app(forall(),p),nil()) -> true() app(app(forall(),p),app(app(cons(),x),xs)) -> app(app(and(),app(p,x)),app(app(forall(),p),xs)) app(app(forsome(),p),nil()) -> false() app(app(forsome(),p),app(app(cons(),x),xs)) -> app(app(or(),app(p,x)),app(app(forsome(),p),xs)) Instead we prove termination of the uncurried system: app(and_0(),y) -> and_1(y) app(and_1(x1),y) -> and_2(x1,y) app(cons_0(),y) -> cons_1(y) app(cons_1(x1),y) -> cons_2(x1,y) app(forall_0(),y) -> forall_1(y) app(forall_1(x1),y) -> forall_2(x1,y) app(forsome_0(),y) -> forsome_1(y) app(forsome_1(x1),y) -> forsome_2(x1,y) app(or_0(),y) -> or_1(y) app(or_1(x1),y) -> or_2(x1,y) and_2(true_0(),true_0()) -> true_0() and_2(x,false_0()) -> false_0() and_2(false_0(),y) -> false_0() or_2(true_0(),y) -> true_0() or_2(x,true_0()) -> true_0() or_2(false_0(),false_0()) -> false_0() forall_2(p,nil_0()) -> true_0() forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) forsome_2(p,nil_0()) -> false_0() forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: app#(and_1(x1),y) -> and_2#(x1,y) p2: app#(forall_1(x1),y) -> forall_2#(x1,y) p3: app#(forsome_1(x1),y) -> forsome_2#(x1,y) p4: app#(or_1(x1),y) -> or_2#(x1,y) p5: forall_2#(p,cons_2(x,xs)) -> and_2#(app(p,x),forall_2(p,xs)) p6: forall_2#(p,cons_2(x,xs)) -> app#(p,x) p7: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs) p8: forsome_2#(p,cons_2(x,xs)) -> or_2#(app(p,x),forsome_2(p,xs)) p9: forsome_2#(p,cons_2(x,xs)) -> app#(p,x) p10: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The estimated dependency graph contains the following SCCs: {p2, p3, p6, p7, p9, p10} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: app#(forall_1(x1),y) -> forall_2#(x1,y) p2: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs) p3: forall_2#(p,cons_2(x,xs)) -> app#(p,x) p4: app#(forsome_1(x1),y) -> forsome_2#(x1,y) p5: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) p6: forsome_2#(p,cons_2(x,xs)) -> app#(p,x) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 8, x2 - 8} forall_1_A(x1) = x1 + 7 forall_2#_A(x1,x2) = max{0, x1 - 8, x2 - 10} cons_2_A(x1,x2) = max{9, x1 + 2, x2} forsome_1_A(x1) = max{7, x1} forsome_2#_A(x1,x2) = max{0, x1 - 8, x2 - 10} app#!_A(x1,x2) = max{7, x1 - 33, x2 - 24} forall_1!_A(x1) = max{8, x1 - 25} forall_2#!_A(x1,x2) = max{7, x1 - 26, x2 - 26} cons_2!_A(x1,x2) = max{0, x1 - 10, x2 - 26} forsome_1!_A(x1) = max{6, x1 - 34} forsome_2#!_A(x1,x2) = max{7, x1 - 33, x2 - 26} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 3, x2 - 4} forall_1_A(x1) = x1 + 2 forall_2#_A(x1,x2) = max{0, x1 - 3, x2 - 4} cons_2_A(x1,x2) = max{x1 + 3, x2} forsome_1_A(x1) = max{2, x1} forsome_2#_A(x1,x2) = max{0, x1 - 3, x2 - 4} app#!_A(x1,x2) = max{4, x1 - 25, x2 - 19} forall_1!_A(x1) = max{3, x1 - 24} forall_2#!_A(x1,x2) = max{4, x1 - 24, x2 - 20} cons_2!_A(x1,x2) = max{1, x1 - 28, x2 - 20} forsome_1!_A(x1) = max{3, x1 - 26} forsome_2#!_A(x1,x2) = max{4, x1 - 25, x2 - 21} partial status: pi(app#) = [1] pi(forall_1) = [1] pi(forall_2#) = [1] pi(cons_2) = [2] pi(forsome_1) = [1] pi(forsome_2#) = [1] The following rules are removed: p1 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs) p2: forall_2#(p,cons_2(x,xs)) -> app#(p,x) p3: app#(forsome_1(x1),y) -> forsome_2#(x1,y) p4: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) p5: forsome_2#(p,cons_2(x,xs)) -> app#(p,x) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The estimated dependency graph contains the following SCCs: {p1} {p3, p4, p5} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: forall_2#(p,cons_2(x,xs)) -> forall_2#(p,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: forall_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = max{x1 - 12, x2 + 1} forall_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 2} cons_2!_A(x1,x2) = max{1, x1 - 42, x2 - 42} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: forall_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = max{1, x1 - 18, x2} forall_2#!_A(x1,x2) = max{7, x1 - 1, x2 - 2} cons_2!_A(x1,x2) = max{x1 - 22, x2 + 6} partial status: pi(forall_2#) = [2] pi(cons_2) = [2] 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: app#(forsome_1(x1),y) -> forsome_2#(x1,y) p2: forsome_2#(p,cons_2(x,xs)) -> app#(p,x) p3: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 15, x2 - 1} forsome_1_A(x1) = x1 + 14 forsome_2#_A(x1,x2) = max{0, x1 - 15, x2 - 10} cons_2_A(x1,x2) = max{x1 + 9, x2} app#!_A(x1,x2) = max{3, x1 - 10, x2 - 1} forsome_1!_A(x1) = max{0, x1 - 2} forsome_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 10} cons_2!_A(x1,x2) = max{2, x1 - 2, x2 + 1} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: app#_A(x1,x2) = max{0, x1 - 6, x2 - 3} forsome_1_A(x1) = max{5, x1 + 3} forsome_2#_A(x1,x2) = max{0, x1 - 6, x2 - 3} cons_2_A(x1,x2) = max{x1 + 2, x2} app#!_A(x1,x2) = max{1, x1 - 2, x2 - 41} forsome_1!_A(x1) = max{1, x1 - 45} forsome_2#!_A(x1,x2) = max{2, x1 - 1, x2 - 42} cons_2!_A(x1,x2) = max{0, x1 - 28, x2 - 41} partial status: pi(app#) = [] pi(forsome_1) = [1] pi(forsome_2#) = [] pi(cons_2) = [2] The following rules are removed: p1, p2 -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The estimated dependency graph contains the following SCCs: {p1} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: forsome_2#(p,cons_2(x,xs)) -> forsome_2#(p,xs) and R consists of: r1: app(and_0(),y) -> and_1(y) r2: app(and_1(x1),y) -> and_2(x1,y) r3: app(cons_0(),y) -> cons_1(y) r4: app(cons_1(x1),y) -> cons_2(x1,y) r5: app(forall_0(),y) -> forall_1(y) r6: app(forall_1(x1),y) -> forall_2(x1,y) r7: app(forsome_0(),y) -> forsome_1(y) r8: app(forsome_1(x1),y) -> forsome_2(x1,y) r9: app(or_0(),y) -> or_1(y) r10: app(or_1(x1),y) -> or_2(x1,y) r11: and_2(true_0(),true_0()) -> true_0() r12: and_2(x,false_0()) -> false_0() r13: and_2(false_0(),y) -> false_0() r14: or_2(true_0(),y) -> true_0() r15: or_2(x,true_0()) -> true_0() r16: or_2(false_0(),false_0()) -> false_0() r17: forall_2(p,nil_0()) -> true_0() r18: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r19: forsome_2(p,nil_0()) -> false_0() r20: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The set of usable rules consists of (no rules) Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: trivial reduction triple (T x T, T x T, Φ) base order 2: lexicographic combination of reduction triples: 1. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: forsome_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = max{1, x1 - 6, x2} forsome_2#!_A(x1,x2) = max{4, x1 - 1, x2 - 2} cons_2!_A(x1,x2) = max{3, x1 - 3, x2 - 4} 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: forsome_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = max{3, x1 - 11, x2} forsome_2#!_A(x1,x2) = max{3, x1 - 1, x2 - 37} cons_2!_A(x1,x2) = max{2, x1 - 28, x2 - 37} partial status: pi(forsome_2#) = [2] pi(cons_2) = [2] The following rules are removed: p1 Then no dependency pair remains.