YES We show the termination of the TRS R: app(app(and(),true()),true()) -> true() app(app(and(),true()),false()) -> false() app(app(and(),false()),true()) -> false() app(app(and(),false()),false()) -> false() app(app(or(),true()),true()) -> true() app(app(or(),true()),false()) -> true() app(app(or(),false()),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(true_0(),false_0()) -> false_0() and_2(false_0(),true_0()) -> false_0() and_2(false_0(),false_0()) -> false_0() or_2(true_0(),true_0()) -> true_0() or_2(true_0(),false_0()) -> true_0() or_2(false_0(),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(true_0(),false_0()) -> false_0() r13: and_2(false_0(),true_0()) -> false_0() r14: and_2(false_0(),false_0()) -> false_0() r15: or_2(true_0(),true_0()) -> true_0() r16: or_2(true_0(),false_0()) -> true_0() r17: or_2(false_0(),true_0()) -> true_0() r18: or_2(false_0(),false_0()) -> false_0() r19: forall_2(p,nil_0()) -> true_0() r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r21: forsome_2(p,nil_0()) -> false_0() r22: 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(true_0(),false_0()) -> false_0() r13: and_2(false_0(),true_0()) -> false_0() r14: and_2(false_0(),false_0()) -> false_0() r15: or_2(true_0(),true_0()) -> true_0() r16: or_2(true_0(),false_0()) -> true_0() r17: or_2(false_0(),true_0()) -> true_0() r18: or_2(false_0(),false_0()) -> false_0() r19: forall_2(p,nil_0()) -> true_0() r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r21: forsome_2(p,nil_0()) -> false_0() r22: forsome_2(p,cons_2(x,xs)) -> or_2(app(p,x),forsome_2(p,xs)) The set of usable rules consists of r12, r13, r14 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 + x2 forall_1_A(x1) = x1 + 2 forall_2#_A(x1,x2) = x1 + x2 cons_2_A(x1,x2) = x1 + x2 + 2 forsome_1_A(x1) = x1 + 2 forsome_2#_A(x1,x2) = x1 + x2 and_2_A(x1,x2) = 4 true_0_A() = 5 false_0_A() = 4 app#!_A(x1,x2) = 0 forall_1!_A(x1) = x1 + 8 forall_2#!_A(x1,x2) = x1 + 3 cons_2!_A(x1,x2) = 2 forsome_1!_A(x1) = 8 forsome_2#!_A(x1,x2) = 3 and_2!_A(x1,x2) = 6 true_0!_A() = 0 false_0!_A() = 5 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: app#_A(x1,x2) = x1 + x2 forall_1_A(x1) = x1 + 3 forall_2#_A(x1,x2) = x1 + x2 cons_2_A(x1,x2) = x1 + x2 + 5 forsome_1_A(x1) = x1 + 3 forsome_2#_A(x1,x2) = x1 + x2 and_2_A(x1,x2) = 7 true_0_A() = 12 false_0_A() = 7 app#!_A(x1,x2) = 0 forall_1!_A(x1) = 8 forall_2#!_A(x1,x2) = x1 + x2 + 9 cons_2!_A(x1,x2) = 3 forsome_1!_A(x1) = x1 + 5 forsome_2#!_A(x1,x2) = x1 + 4 and_2!_A(x1,x2) = x1 + 1 true_0!_A() = 11 false_0!_A() = 12 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] pi(and_2) = [] pi(true_0) = [] pi(false_0) = [] The following rules are removed: p1, p2, p3, p4, p6 -- 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(true_0(),false_0()) -> false_0() r13: and_2(false_0(),true_0()) -> false_0() r14: and_2(false_0(),false_0()) -> false_0() r15: or_2(true_0(),true_0()) -> true_0() r16: or_2(true_0(),false_0()) -> true_0() r17: or_2(false_0(),true_0()) -> true_0() r18: or_2(false_0(),false_0()) -> false_0() r19: forall_2(p,nil_0()) -> true_0() r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r21: forsome_2(p,nil_0()) -> false_0() r22: 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(true_0(),false_0()) -> false_0() r13: and_2(false_0(),true_0()) -> false_0() r14: and_2(false_0(),false_0()) -> false_0() r15: or_2(true_0(),true_0()) -> true_0() r16: or_2(true_0(),false_0()) -> true_0() r17: or_2(false_0(),true_0()) -> true_0() r18: or_2(false_0(),false_0()) -> false_0() r19: forall_2(p,nil_0()) -> true_0() r20: forall_2(p,cons_2(x,xs)) -> and_2(app(p,x),forall_2(p,xs)) r21: forsome_2(p,nil_0()) -> false_0() r22: 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!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: forsome_2#_A(x1,x2) = x2 cons_2_A(x1,x2) = x2 + 2 forsome_2#!_A(x1,x2) = x2 cons_2!_A(x1,x2) = x1 + x2 + 10 2. marked-algebra reduction triple (>=A, >=A!, >A!): matrix interpretation with standard point-wise order: carrier: N^1 order: standard order interpretations: forsome_2#_A(x1,x2) = 0 cons_2_A(x1,x2) = x1 + 5 forsome_2#!_A(x1,x2) = 0 cons_2!_A(x1,x2) = x1 + 29 partial status: pi(forsome_2#) = [2] pi(cons_2) = [1, 2] The following rules are removed: p1 Then no dependency pair remains.