Main derivation
S01 · definition
Before
\text{fixed step, constant factor }r\ge 0After
x_{n+1}=r\,x_n\ (n\in\mathbb{N}),\ x_0\ \text{given}Conditions: r\ge 0
Rule: mko_exp_model · Uses: G1, G2
Leaf basis: mko_exp_model + def_intro
State the model as in the MKO: each step multiplies the current quantity by the same nonnegative factor r. G1 requires fixed steps and an r that does not change over time; these are assumptions of the model, not conclusions from any data.
S02 · definition
Before
r^n
After
r^0=1,\ r^{n+1}=r^n\cdot rConditions: none
Rule: nat_power_def · Uses: nat_power_def
Leaf basis: nat_power_def + def_intro
Make the meaning of powers precise: r^n is the product of n copies of r, defined formally by recursion. The definition also applies to r=0, so 0^0=1 and the closed form makes sense when r=0.
S03 · implies
Before
x_{n+1}=r\,x_n,\ x_0After
\forall n\in\mathbb{N}:\ x_n=x_0\,r^nConditions: none
Rule: L_closed_form · Uses: S01, S02, L_closed_form
Leaf basis: nat_induction + nat_power_def + mul_assoc + mul_comm
By Lemma L_closed_form (proved by induction in S03.1 to S03.3), the recurrence implies the closed form for all n, not just for the few terms that were computed.
S04 · implies
Before
\forall n:\ x_n=x_0\,r^n
After
x_n=8\left(\tfrac{3}{2}\right)^nConditions: x_0=8,\ r=\tfrac{3}{2}\ge 0
Rule: forall_elim · Uses: S03, G3
Leaf basis: forall_elim
The lemma holds for any x_0 and r≥0; substituting the main instance x_0=8, r=3/2 gives x_n=8·(3/2)^n.
S05 · implies
Before
x_0=8,\ x_{n+1}=\tfrac{3}{2}x_nAfter
x_1=12,\ x_2=18,\ x_3=27,\ x_4=\tfrac{81}{2}Conditions: none
Rule: numeric_evaluation · Uses: G2, G3
Leaf basis: numeric_evaluation + eq_replace
Compute step by step from the recurrence: 8·3/2=12, 12·3/2=18, 18·3/2=27, 27·3/2=81/2.
S06 · equivalent
Before
8\left(\tfrac{3}{2}\right)^4After
8\cdot\tfrac{81}{16}=\tfrac{81}{2}=x_4Conditions: none
Rule: power_laws · Uses: S04, S05
Leaf basis: power_laws + frac_rules + numeric_evaluation
The closed form at n=4: (3/2)^4=81/16 and 8·81/16=81/2, agreeing with the recursive computation of S05. This is a numerical cross-check of S03, not its proof.
S07 · equivalent
Before
y_n=x_0\,r^n
After
y_0=x_0,\ y_{n+1}=r\,y_nConditions: none
Rule: nat_power_def · Uses: S03, S02
Leaf basis: nat_power_def + mul_assoc + mul_comm + iff_intro
The other direction: the closed-form sequence y_n=x_0·r^n satisfies the recurrence with the same initial value, since y_0=x_0·1=x_0 and y_(n+1)=x_0·r^n·r=r·y_n. Together with S03 (any sequence satisfying the recurrence and initial value equals the closed form), the recurrence with initial value and the closed form describe the same sequence, both ways, which is the equivalence in the MKO's second paragraph.
S08 · implies
Before
x_{n+1}-x_n=(r-1)x_n,\ r=\tfrac{3}{2}After
x_{n+1}-x_n=\tfrac{1}{2}x_n>0Conditions: x_n=8(\tfrac{3}{2})^n>0
Rule: mko_exp_cases · Uses: S04, mko_exp_cases
Leaf basis: distributive + order_basics + mko_exp_cases
x_(n+1)−x_n=r·x_n−x_n=(r−1)x_n. In the main instance r−1=1/2, and x_n=8·(3/2)^n is a product of positive numbers, so x_n>0; hence each step strictly increases, and the increment itself is proportional to x_n: this is the 'growth by a factor' of r>1.
S09 · implies
Before
x_0=8,\ r=\tfrac{1}{2}After
x_n=8\left(\tfrac{1}{2}\right)^n:\ 8,4,2,1,\tfrac{1}{2};\ x_{n+1}-x_n=-\tfrac{1}{2}x_n<0Conditions: 0\le r<1
Rule: mko_exp_cases · Uses: S03, G3, mko_exp_cases
Leaf basis: L_closed_form + order_basics + mko_exp_cases
Comparison instance r=1/2: by L_closed_form, x_n=8·(1/2)^n, with first five terms 8, 4, 2, 1, 1/2; x_(n+1)−x_n=(1/2−1)x_n=−x_n/2<0, halving at each step. This is the shrinking case 0≤r<1. The general classification does not add an assumption x0≥0: by the closed form, for r>0, x_n has the sign of x0 and x_(n+1)−x_n=(r−1)x_n. Thus for r>1 the sequence strictly increases when x0>0 and strictly decreases when x0<0; its magnitude grows by the factor r for nonzero x0. For 0<r<1 the magnitude shrinks, with positive initial values decreasing and negative initial values increasing. For r=0, x1 and all later values are 0. The constant cases r=1 and x0=0 are handled in S10.
S10 · implies
Before
r=1\ \text{or}\ x_0=0After
r=1:\ x_n=x_0;\ \ x_0=0:\ x_n=0\cdot r^n=0
Conditions: none
Rule: mko_exp_cases · Uses: S03, mko_exp_cases
Leaf basis: L_closed_form + mul_zero + mul_one
For r=1, 1^n=1, so x_n=x_0 and the sequence is constant. For x_0=0 and any r≥0 (including r=0, with 0^0=1 by definition), x_n=0·r^n=0, so the sequence is all zeros.
S11 · implies
Before
d=(8,12,18,26),\ \text{fixed-}r\ \text{model with }x_0=8After
r=\tfrac{3}{2}\ \text{forced},\ x_3=27\neq 26:\ \text{no fixed-}r\ \text{model fits}Conditions: d_0=8\neq 0
Rule: contradiction · Uses: G4, S03
Leaf basis: contradiction + mul_inverse_nonzero + L_closed_form
Assume some fixed-factor model x_(n+1)=r·x_n with x_0=8 matches all four data. From x_1=12=r·8 and 8≠0, r=12/8=3/2 (S11.1). By L_closed_form the model gives x_3=8·(3/2)^3=27, but d_3=26, a contradiction (S11.2). So no fixed-r model matches all four data. The first three data have ratios 3/2, which looks like a fixed factor, but that does not determine the fourth period.
S12 · implies
Before
r=\tfrac{3}{2}\ \text{per step},\ \text{continuous }x(t)=x_0e^{kt}After
e^{k}=\tfrac{3}{2}\ \text{needed};\ e^{1/2}\ge\tfrac{13}{8}>\tfrac{3}{2},\ \text{so }k\neq\tfrac{1}{2}Conditions: u=\tfrac{1}{2}\ge 0
Rule: exp_lower_bound · Uses: G1, exp_lower_bound
Leaf basis: exp_lower_bound + order_basics
A per-step factor 3/2 means a 50% increase per step. For the worked nonzero initial state x_0=8, agreement of x(t)=x_0·e^(kt) with the model at integer times permits cancellation of x_0 at time 1, giving e^k=3/2. If x_0=0 both trajectories are zero and their agreement does not determine k; the following factor comparison concerns the nonzero trajectory. Taking the 50% directly as the continuous rate k=1/2 is wrong: by the series bound e^(1/2)≥1+1/2+1/8=13/8, and 13/8>3/2, so e^(1/2)≠3/2. A discrete factor and a continuous growth rate are different quantities; no continuous model is built here.
S13 · implies
Before
\text{proved facts}After
x_n=x_0r^n\ \forall n;\ x_4=\tfrac{81}{2};\ \text{data }(8,12,18,26)\ \text{fit no fixed }rConditions: none
Rule: and_intro_elim · Uses: S03, S06, S07, S08, S09, S10, S11, S12
Leaf basis: and_intro_elim
Combine the conclusions: in the model x_n=x_0·r^n for all n, equivalent to the recurrence; in the main instance x_4=81/2; for nonzero initial values the magnitude grows when r>1 and shrinks when 0<r<1, with the numerical monotonicity depending on the initial sign; r=0 gives zero from the first step, r=1 is constant, and x_0=0 stays zero; the data 8, 12, 18, 26 fit no fixed-factor model; the per-step factor is not a continuous rate.
Substeps and lemmas
S03.1 · induction_base
Before
n=0
After
x_0=x_0\cdot 1=x_0\,r^0
Conditions: none
Rule: nat_induction · Uses: S02
Leaf basis: nat_power_def + mul_one
Base: let P(n) be x_n=x_0·r^n. Since r^0=1 and x_0·1=x_0, P(0) holds.
S03.2 · induction_step
Before
x_n=x_0\,r^n
After
x_{n+1}=r\,x_n=r\,(x_0\,r^n)=x_0\,(r^n\cdot r)=x_0\,r^{n+1}Conditions: none
Rule: eq_replace · Uses: S01, S02
Leaf basis: eq_replace + mul_assoc + mul_comm + nat_power_def
Inductive step: assume P(n). By the recurrence x_(n+1)=r·x_n; replace x_n by x_0·r^n, rearrange by commutativity and associativity of multiplication to x_0·(r^n·r), which by the definition of powers is x_0·r^(n+1); so P(n+1) holds.
S03.3 · implies
Before
P(0),\ \forall n\,(P(n)\Rightarrow P(n+1))
After
\forall n\in\mathbb{N}:\ x_n=x_0\,r^nConditions: none
Rule: nat_induction · Uses: S03.1, S03.2
Leaf basis: nat_induction
By the principle of induction, P(n) holds for every natural number n. S03.1 to S03.3 together prove Lemma L_closed_form.
S11.1 · equivalent
Before
12=r\cdot 8
After
r=\tfrac{12}{8}=\tfrac{3}{2}Conditions: 8\neq 0
Rule: mul_inverse_nonzero · Uses: G4, G2
Leaf basis: mul_inverse_nonzero + eq_congruence + frac_rules
Multiply both sides by 1/8 (since 8≠0): r=12/8=3/2.
S11.2 · implies
Before
x_3=8\left(\tfrac{3}{2}\right)^3After
x_3=27\neq 26=d_3
Conditions: none
Rule: numeric_evaluation · Uses: S11.1, S03, G4
Leaf basis: numeric_evaluation + power_laws
8·27/8=27, while d_3=26; the numbers differ, contradicting the assumption that the model fits the data. Also d_3/d_2=26/18=13/9≠3/2, showing that the ratio at the third step differs.