The Runtime Complexity (full) of the given CpxTRS could be proven to be BOUNDS(n^1, INF).

(0) Obligation:

Runtime Complexity TRS:
The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0), gr(y, 0)), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0, 0) → false
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Rewrite Strategy: FULL

(1) DecreasingLoopProof (EQUIVALENT transformation)

The following loop(s) give(s) rise to the lower bound Ω(n¹):
The rewrite sequence
cond(true, s(x15_1), s(x168_1)) →⁺ cond(true, x15_1, x168_1)
gives rise to a decreasing loop by considering the right hand sides subterm at position [].
The pumping substitution is [x15_1 / s(x15_1), x168_1 / s(x168_1)].
The result substitution is [ ].

(2) BOUNDS(n^1, INF)

(3) RenamingProof (EQUIVALENT transformation)

Renamed function symbols to avoid clashes with predefined symbol.

(4) Obligation:

Runtime Complexity Relative TRS:
The TRS R consists of the following rules:

cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

S is empty.
Rewrite Strategy: FULL

(5) TypeInferenceProof (BOTH BOUNDS(ID, ID) transformation)

Infered types.

(6) Obligation:

TRS:
Rules:
cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

Types:
cond :: true:false → 0':s → 0':s → cond
true :: true:false
and :: true:false → true:false → true:false
gr :: 0':s → 0':s → true:false
0' :: 0':s
p :: 0':s → 0':s
false :: true:false
s :: 0':s → 0':s
hole_cond1_0 :: cond
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
gen_0':s4_0 :: Nat → 0':s

(7) OrderProof (LOWER BOUND(ID) transformation)

Heuristically decided to analyse the following defined symbols:
cond, gr

They will be analysed ascendingly in the following order:
gr < cond

(8) Obligation:

TRS:
Rules:
cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

Types:
cond :: true:false → 0':s → 0':s → cond
true :: true:false
and :: true:false → true:false → true:false
gr :: 0':s → 0':s → true:false
0' :: 0':s
p :: 0':s → 0':s
false :: true:false
s :: 0':s → 0':s
hole_cond1_0 :: cond
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
gen_0':s4_0 :: Nat → 0':s

Generator Equations:
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

The following defined symbols remain to be analysed:
gr, cond

They will be analysed ascendingly in the following order:
gr < cond

(9) RewriteLemmaProof (LOWER BOUND(ID) transformation)

Proved the following rewrite lemma:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

Induction Base:
gr(gen_0':s4_0(0), gen_0':s4_0(0)) →_R^Ω(1)
false

Induction Step:
gr(gen_0':s4_0(+(n6_0, 1)), gen_0':s4_0(+(n6_0, 1))) →_R^Ω(1)
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) →_IH
false

We have rt ∈ Ω(n¹) and sz ∈ O(n). Thus, we have irc_R ∈ Ω(n).

(10) Complex Obligation (BEST)

(11) Obligation:

TRS:
Rules:
cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

Types:
cond :: true:false → 0':s → 0':s → cond
true :: true:false
and :: true:false → true:false → true:false
gr :: 0':s → 0':s → true:false
0' :: 0':s
p :: 0':s → 0':s
false :: true:false
s :: 0':s → 0':s
hole_cond1_0 :: cond
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
gen_0':s4_0 :: Nat → 0':s

Lemmas:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

The following defined symbols remain to be analysed:
cond

(12) NoRewriteLemmaProof (LOWER BOUND(ID) transformation)

Could not prove a rewrite lemma for the defined symbol cond.

(13) Obligation:

TRS:
Rules:
cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

Types:
cond :: true:false → 0':s → 0':s → cond
true :: true:false
and :: true:false → true:false → true:false
gr :: 0':s → 0':s → true:false
0' :: 0':s
p :: 0':s → 0':s
false :: true:false
s :: 0':s → 0':s
hole_cond1_0 :: cond
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
gen_0':s4_0 :: Nat → 0':s

Lemmas:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

No more defined symbols left to analyse.

(14) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

(15) BOUNDS(n^1, INF)

(16) Obligation:

TRS:
Rules:
cond(true, x, y) → cond(and(gr(x, 0'), gr(y, 0')), p(x), p(y))
and(true, true) → true
and(x, false) → false
and(false, x) → false
gr(0', 0') → false
gr(0', x) → false
gr(s(x), 0') → true
gr(s(x), s(y)) → gr(x, y)
p(0') → 0'
p(s(x)) → x

Types:
cond :: true:false → 0':s → 0':s → cond
true :: true:false
and :: true:false → true:false → true:false
gr :: 0':s → 0':s → true:false
0' :: 0':s
p :: 0':s → 0':s
false :: true:false
s :: 0':s → 0':s
hole_cond1_0 :: cond
hole_true:false2_0 :: true:false
hole_0':s3_0 :: 0':s
gen_0':s4_0 :: Nat → 0':s

Lemmas:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

Generator Equations:
gen_0':s4_0(0) ⇔ 0'
gen_0':s4_0(+(x, 1)) ⇔ s(gen_0':s4_0(x))

No more defined symbols left to analyse.

(17) LowerBoundsProof (EQUIVALENT transformation)

The lowerbound Ω(n¹) was proven with the following lemma:
gr(gen_0':s4_0(n6_0), gen_0':s4_0(n6_0)) → false, rt ∈ Ω(1 + n6₀)

(18) BOUNDS(n^1, INF)