Main derivation
S01 · definition
Before
\text{discrete logistic increment model}After
G(x)=x+rx\left(1-\tfrac{x}{K}\right),\ K>0,\ 0\le r\le 1Conditions: K>0; 0\le r\le 1
Rule: mko_logistic · Uses: G1, mko_logistic
Leaf basis: mko_logistic + def_intro
Fix the model specified by the MKO: the next state is the current value plus the increment r x(1−x/K). This is one particular discrete model; other maps also called 'logistic' are different models, and the conclusions here do not transfer to them (see the error controls).
S02 · equivalent
Before
G(x)-x
After
\tfrac{rx(K-x)}{K}Conditions: K\neq 0
Rule: frac_rules · Uses: S01, frac_rules, distributive
Leaf basis: frac_rules + distributive + mul_assoc
The increment is G(x)−x=r x(1−x/K). Since K≠0, 1−x/K=K/K−x/K=(K−x)/K, so the increment equals r x(K−x)/K. This writes the increment as the product of the three factors r, x, K−x divided by the positive number K, which makes its sign easy to read.
S03 · implies
Before
0\le x\le K,\ 0\le r,\ K>0
After
G(x)-x=\tfrac{rx(K-x)}{K}\ge 0Conditions: 0\le x\le K; r\ge 0; K>0
Rule: L_nonneg_prod · Uses: S02, G1
Leaf basis: L_nonneg_prod + order_basics
For 0≤x≤K the three factors r≥0, x≥0 and K−x≥0 are all nonnegative. Applying Lemma L_nonneg_prod twice gives r x(K−x)≥0; multiplying by the positive number 1/K keeps the direction, so the increment is ≥0.
S04 · implies
Before
G(x)-x\ge 0,\ x\ge 0
After
G(x)\ge x\ge 0
Conditions: none
Rule: order_basics · Uses: S03, G1
Leaf basis: order_basics
Adding x to both sides of G(x)−x≥0 gives G(x)≥x; with x≥0, transitivity gives G(x)≥0. The lower bound is proved, and it also shows that on [0,K] an update never decreases the value.
S05 · equivalent
Before
K\cdot(K-G(x))
After
(K-x)(K-rx)
Conditions: K>0
Rule: L_gap · Uses: S01
Leaf basis: distributive + frac_rules
The key identity for the upper bound (Lemma L_gap): the gap between capacity and the next state, multiplied by K, factors exactly as (K−x)(K−r x). This turns 'G(x)≤K' into a question about the signs of two factors.
S06 · implies
Before
0\le r\le 1,\ 0\le x\le K
After
rx\le x\le K
Conditions: r\le 1; x\ge 0
Rule: order_basics · Uses: G1, order_basics
Leaf basis: order_basics + distributive + L_nonneg_prod
x−r x=(1−r)x, where 1−r≥0 (since r≤1) and x≥0, so by L_nonneg_prod (1−r)x≥0, that is r x≤x; with x≤K this gives r x≤K. This is the only place in the proof where r≤1 is used.
S07 · implies
Before
K-x\ge 0,\ K-rx\ge 0,\ K>0
After
G(x)\le K
Conditions: K>0
Rule: L_nonneg_prod · Uses: S05, S06, L_nonneg_prod
Leaf basis: L_nonneg_prod + order_basics + L_gap
K−x≥0 (since x≤K) and K−r x≥0 (S06), so by L_nonneg_prod (K−x)(K−r x)≥0; by S05, K(K−G(x))≥0. Since K>0, multiplying by 1/K gives K−G(x)≥0, that is G(x)≤K. The upper bound is proved.
S08 · implies
Before
0\le x\le K
After
0\le G(x)\le K
Conditions: K>0; 0\le r\le 1
Rule: and_intro_elim · Uses: S04, S07
Leaf basis: and_intro_elim
Combine the lower bound (S04) and the upper bound (S07): one-step invariance holds. This is exactly the MKO's statement 'if 0≤x_n≤K, the next state remains in [0,K]', and the proof holds uniformly for every x in the interval; it is not a finite check.
S09 · implies
Before
x_0\in[0,K]
After
\forall n\in\mathbb{N}:\ x_n\in[0,K]Conditions: none
Rule: nat_induction · Uses: S08, G1, nat_induction
Leaf basis: nat_induction + modus_ponens
Induction on n: the base x0∈[0,K] is given by G1; if x_n∈[0,K], then by S08 x_(n+1)=G(x_n)∈[0,K]. So the whole orbit stays in [0,K].
S10 · equivalent
Before
G(x)=x
After
rx(K-x)=0;\ G(0)=0,\ G(K)=K;\ r>0\Rightarrow x\in\{0,K\}Conditions: K\neq 0
Rule: zero_product · Uses: S02, G1, zero_product
Leaf basis: zero_product + mul_inverse_nonzero + case_analysis
By S02, G(x)=x ⇔ r x(K−x)/K=0 ⇔ r x(K−x)=0 (multiplying or dividing by K≠0). x=0 and x=K make the product zero, so 0 and K are fixed points for every r. If r>0, the zero-product property gives r x(K−x)=0 ⇔ x=0 or x=K, so they are the only fixed points. If r=0, G(x)=x for every x and every point is fixed.
S11 · equivalent
Before
x_0=2,\ K=8,\ r=1
After
x_1=2+2\left(1-\tfrac{2}{8}\right)=\tfrac{7}{2}Conditions: none
Rule: numeric_evaluation · Uses: G2, S01
Leaf basis: numeric_evaluation + frac_rules
With r=1, G(x)=x+x(1−x/8)=x(2−x/8). x1=2·(2−1/4)=2·7/4=7/2, all in exact fractions (N3).
S12 · equivalent
Before
x_1=\tfrac{7}{2}After
x_2=\tfrac{7}{2}\left(2-\tfrac{7}{16}\right)=\tfrac{7}{2}\cdot\tfrac{25}{16}=\tfrac{175}{32}Conditions: none
Rule: numeric_evaluation · Uses: S11
Leaf basis: numeric_evaluation + frac_rules
x1/8=7/16, 2−7/16=25/16, and x2=(7/2)(25/16)=175/32 (N4).
S13 · implies
Before
x_2=\tfrac{175}{32}After
x_3=\tfrac{175}{32}\cdot\tfrac{337}{256}=\tfrac{58975}{8192};\ 2<\tfrac{7}{2}<\tfrac{175}{32}<\tfrac{58975}{8192}<8Conditions: none
Rule: numeric_evaluation · Uses: S12, S09, S04
Leaf basis: numeric_evaluation + frac_rules + order_basics
x2/8=175/256, 2−175/256=337/256, and x3=(175/32)(337/256)=58975/8192 (N5, N6), about 7.199 and below 8 (N7). All four values lie in [0,8] and increase, consistent with S09 (staying in the interval) and S04 (G(x)≥x). These three steps are only an example of consistency; they prove neither that the orbit converges to 8 nor that 8 is stable (MKO assumption a2).
S14 · implies
Before
K=8,\ r=3,\ x=6
After
G(6)=6+3\cdot 6\cdot\tfrac{1}{4}=\tfrac{21}{2}>8Conditions: r=3>1
Rule: numeric_evaluation · Uses: G3, S05
Leaf basis: numeric_evaluation + L_gap
The condition r≤1 cannot be dropped: with r=3, r x=18>8=K and S06 fails; L_gap gives 8(8−G(6))=(8−6)(8−18)=−20, so G(6)=8+20/8=21/2>8, which leaves [0,8]. This concrete counterexample shows that S08 depends on r≤1.
S15 · implies
Before
\text{model with}\ K>0,\ 0\le r\le 1After
G([0,K])\subseteq[0,K],\ \mathrm{Fix}\supseteq\{0,K\}\ (=\{0,K\}\ \text{if}\ r>0)Conditions: none
Rule: and_intro_elim · Uses: S08, S09, S10, S13, S14
Leaf basis: and_intro_elim
Summary: [0,K] is invariant under G (S08), so an orbit starting in [0,K] stays there forever (S09); 0 and K are fixed points, and the only ones when r>0 (S10); the example orbit 2, 7/2, 175/32, 58975/8192 agrees with these conclusions (S13); and r≤1 cannot be dropped (S14).