YES We show the termination of the TRS R: half(|0|()) -> |0|() half(s(|0|())) -> |0|() half(s(s(x))) -> s(half(x)) lastbit(|0|()) -> |0|() lastbit(s(|0|())) -> s(|0|()) lastbit(s(s(x))) -> lastbit(x) zero(|0|()) -> true() zero(s(x)) -> false() conv(x) -> conviter(x,cons(|0|(),nil())) conviter(x,l) -> if(zero(x),x,l) if(true(),x,l) -> l if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l)) -- SCC decomposition. Consider the dependency pair problem (P, R), where P consists of p1: half#(s(s(x))) -> half#(x) p2: lastbit#(s(s(x))) -> lastbit#(x) p3: conv#(x) -> conviter#(x,cons(|0|(),nil())) p4: conviter#(x,l) -> if#(zero(x),x,l) p5: conviter#(x,l) -> zero#(x) p6: if#(false(),x,l) -> conviter#(half(x),cons(lastbit(x),l)) p7: if#(false(),x,l) -> half#(x) p8: if#(false(),x,l) -> lastbit#(x) and R consists of: r1: half(|0|()) -> |0|() r2: half(s(|0|())) -> |0|() r3: half(s(s(x))) -> s(half(x)) r4: lastbit(|0|()) -> |0|() r5: lastbit(s(|0|())) -> s(|0|()) r6: lastbit(s(s(x))) -> lastbit(x) r7: zero(|0|()) -> true() r8: zero(s(x)) -> false() r9: conv(x) -> conviter(x,cons(|0|(),nil())) r10: conviter(x,l) -> if(zero(x),x,l) r11: if(true(),x,l) -> l r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l)) The estimated dependency graph contains the following SCCs: {p4, p6} {p1} {p2} -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: conviter#(x,l) -> if#(zero(x),x,l) p2: if#(false(),x,l) -> conviter#(half(x),cons(lastbit(x),l)) and R consists of: r1: half(|0|()) -> |0|() r2: half(s(|0|())) -> |0|() r3: half(s(s(x))) -> s(half(x)) r4: lastbit(|0|()) -> |0|() r5: lastbit(s(|0|())) -> s(|0|()) r6: lastbit(s(s(x))) -> lastbit(x) r7: zero(|0|()) -> true() r8: zero(s(x)) -> false() r9: conv(x) -> conviter(x,cons(|0|(),nil())) r10: conviter(x,l) -> if(zero(x),x,l) r11: if(true(),x,l) -> l r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l)) The set of usable rules consists of r1, r2, r3, r4, r5, r6, r7, r8 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: conviter#_A(x1,x2) = x2 if#_A(x1,x2,x3) = x3 zero_A(x1) = x1 + 1 false_A() = 7 half_A(x1) = 6 cons_A(x1,x2) = x2 lastbit_A(x1) = 6 |0|_A() = 0 s_A(x1) = 6 true_A() = 1 conviter#!_A(x1,x2) = 1 if#!_A(x1,x2,x3) = 1 zero!_A(x1) = x1 false!_A() = 5 half!_A(x1) = 6 cons!_A(x1,x2) = 2 lastbit!_A(x1) = 7 |0|!_A() = 2 s!_A(x1) = 6 true!_A() = 1 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: conviter#_A(x1,x2) = max{6, x2 + 5} if#_A(x1,x2,x3) = max{6, x3 - 1} zero_A(x1) = max{0, x1 - 4} false_A = 13 half_A(x1) = max{11, x1 - 2} cons_A(x1,x2) = max{0, x1 - 21} lastbit_A(x1) = 17 |0|_A = 3 s_A(x1) = max{17, x1 + 5} true_A = 0 conviter#!_A(x1,x2) = max{x1 + 6, x2 + 3} if#!_A(x1,x2,x3) = max{x1 + 5, x2 + 5, x3 + 2} zero!_A(x1) = max{11, x1 - 18} false!_A = 2 half!_A(x1) = x1 + 5 cons!_A(x1,x2) = max{0, x1 - 17, x2 - 21} lastbit!_A(x1) = 23 |0|!_A = 8 s!_A(x1) = x1 + 5 true!_A = 7 partial status: pi(conviter#) = [] pi(if#) = [] pi(zero) = [1] pi(false) = [] pi(half) = [] pi(cons) = [] pi(lastbit) = [] pi(|0|) = [] pi(s) = [] pi(true) = [] The following rules are removed: p1, p2 Then no dependency pair remains. -- Reduction pair. Consider the dependency pair problem (P, R), where P consists of p1: half#(s(s(x))) -> half#(x) and R consists of: r1: half(|0|()) -> |0|() r2: half(s(|0|())) -> |0|() r3: half(s(s(x))) -> s(half(x)) r4: lastbit(|0|()) -> |0|() r5: lastbit(s(|0|())) -> s(|0|()) r6: lastbit(s(s(x))) -> lastbit(x) r7: zero(|0|()) -> true() r8: zero(s(x)) -> false() r9: conv(x) -> conviter(x,cons(|0|(),nil())) r10: conviter(x,l) -> if(zero(x),x,l) r11: if(true(),x,l) -> l r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l)) 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: half#_A(x1) = x1 s_A(x1) = x1 + 1 half#!_A(x1) = x1 s!_A(x1) = 5 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: half#_A(x1) = x1 s_A(x1) = max{53, x1 + 17} half#!_A(x1) = max{45, x1 - 1} s!_A(x1) = max{7, x1 - 9} partial status: pi(half#) = [1] pi(s) = [1] 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: lastbit#(s(s(x))) -> lastbit#(x) and R consists of: r1: half(|0|()) -> |0|() r2: half(s(|0|())) -> |0|() r3: half(s(s(x))) -> s(half(x)) r4: lastbit(|0|()) -> |0|() r5: lastbit(s(|0|())) -> s(|0|()) r6: lastbit(s(s(x))) -> lastbit(x) r7: zero(|0|()) -> true() r8: zero(s(x)) -> false() r9: conv(x) -> conviter(x,cons(|0|(),nil())) r10: conviter(x,l) -> if(zero(x),x,l) r11: if(true(),x,l) -> l r12: if(false(),x,l) -> conviter(half(x),cons(lastbit(x),l)) The set of usable rules consists of r4 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: lastbit#_A(x1) = x1 s_A(x1) = x1 + 1 lastbit_A(x1) = x1 + 1 |0|_A() = 3 lastbit#!_A(x1) = 5 s!_A(x1) = 6 lastbit!_A(x1) = 6 |0|!_A() = 1 2. marked-algebra reduction triple (>=A, >=A!, >A!): max/plus interpretation on natural numbers: lastbit#_A(x1) = 0 s_A(x1) = max{19, x1 - 6} lastbit_A(x1) = max{0, x1 - 26} |0|_A = 0 lastbit#!_A(x1) = 15 s!_A(x1) = max{20, x1 + 14} lastbit!_A(x1) = max{9, x1 + 8} |0|!_A = 3 partial status: pi(lastbit#) = [1] pi(s) = [1] pi(lastbit) = [1] pi(|0|) = [] The following rules are removed: p1 Then no dependency pair remains.