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, r7, r8 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: conviter#_A(x1,x2) = max{24, x1 + 5, x2 + 13} if#_A(x1,x2,x3) = max{12, x1 - 13, x2, x3 - 1} zero_A(x1) = max{25, x1 + 8} false_A = 37 half_A(x1) = max{19, x1 - 5} cons_A(x1,x2) = 11 lastbit_A(x1) = max{87, x1 + 9} |0|_A = 18 s_A(x1) = max{56, x1 + 30} true_A = 17 base order 2: 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) = 0 if#_A(x1,x2,x3) = 0 zero_A(x1) = x1 + 1 false_A() = 7 half_A(x1) = 6 cons_A(x1,x2) = x2 + 1 lastbit_A(x1) = 1 |0|_A() = 0 s_A(x1) = 6 true_A() = 1 conviter#!_A(x1,x2) = 2 if#!_A(x1,x2,x3) = x1 + x3 zero!_A(x1) = 3 false!_A() = 12 half!_A(x1) = 4 cons!_A(x1,x2) = 7 lastbit!_A(x1) = x1 + 6 |0|!_A() = 3 s!_A(x1) = 2 true!_A() = 0 partial status: pi(conviter#) = [] pi(if#) = [2] pi(zero) = [] pi(false) = [] pi(half) = [] pi(cons) = [] pi(lastbit) = [] pi(|0|) = [] pi(s) = [1] 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 r4, r5 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: half#_A(x1) = max{30, x1 + 20} s_A(x1) = max{19, x1 + 10} lastbit_A(x1) = max{19, x1 + 10} |0|_A = 10 base order 2: 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 lastbit_A(x1) = x1 + 1 |0|_A() = 3 half#!_A(x1) = x1 + 9 s!_A(x1) = x1 + 7 lastbit!_A(x1) = x1 + 5 |0|!_A() = 2 partial status: pi(half#) = [1] pi(s) = [1] pi(lastbit) = [] pi(|0|) = [] 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, r5 Take the reduction pair: generalized weighted path order (>=MG, >G) base order 1: max/plus interpretation on natural numbers: lastbit#_A(x1) = max{30, x1 + 20} s_A(x1) = max{19, x1 + 10} lastbit_A(x1) = max{19, x1 + 10} |0|_A = 10 base order 2: 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 + 2 lastbit_A(x1) = x1 + 1 |0|_A() = 4 lastbit#!_A(x1) = x1 + 10 s!_A(x1) = x1 + 7 lastbit!_A(x1) = x1 + 4 |0|!_A() = 9 partial status: pi(lastbit#) = [1] pi(s) = [1] pi(lastbit) = [] pi(|0|) = [] The following rules are removed: p1 Then no dependency pair remains.