(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Combined order from the following AFS and order.
terms(x1)  =  terms(x1)
cons(x1, x2)  =  cons(x1, x2)
recip(x1)  =  x1
sqr(x1)  =  sqr(x1)
n__terms(x1)  =  x1
s(x1)  =  s(x1)
0  =  0
add(x1, x2)  =  add(x1, x2)
dbl(x1)  =  dbl(x1)
first(x1, x2)  =  first(x1, x2)
nil  =  nil
n__first(x1, x2)  =  n__first(x1, x2)
activate(x1)  =  activate(x1)
half(x1)  =  half(x1)

Recursive path order with status [RPO].
Quasi-Precedence:

[terms₁, first₂, activate₁] > cons₂ > n_first2
[terms₁, first₂, activate₁] > sqr₁ > add₂ > [s₁, half₁] > 0
[terms₁, first₂, activate₁] > sqr₁ > dbl₁ > [s₁, half₁] > 0
[terms₁, first₂, activate₁] > nil

Status:

terms₁: multiset
cons₂: multiset
sqr₁: [1]
s₁: multiset
0: multiset
add₂: multiset
dbl₁: multiset
first₂: multiset
nil: multiset
n_first2: multiset
activate₁: multiset
half₁: multiset

With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

terms(N) → cons(recip(sqr(N)), n__terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, n__first(X, activate(Z)))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X
terms(X) → n__terms(X)
first(X1, X2) → n__first(X1, X2)
activate(n__first(X1, X2)) → first(X1, X2)
activate(X) → X

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:

activate₁ > terms₁ > n_terms1

and weight map:

activate_1=1
n__terms_1=1
terms_1=2

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

activate(n__terms(X)) → terms(X)

(5) RisEmptyProof (EQUIVALENT transformation)

The TRS R is empty. Hence, termination is trivially proven.