Main derivation
S01 · equivalent
Before
c(101)
After
\left(\tfrac13\right)^2\left(\tfrac23\right)^1=\tfrac{2}{27}Conditions: none
Rule: numeric_evaluation · Uses: G1, G2
Leaf basis: numeric_evaluation + power_laws
First read the certificate: w=101 has length n=3 and k=2 ones, so c(101)=p^2(1−p)^1=(1/3)^2(2/3)=2/27 (N1). The formula depends only on the number of ones, so 110 and 011 also get 2/27 (N9).
S02 · equivalent
Before
c(w)=p^{k(w)}(1-p)^{n-k(w)}After
c(w)=\prod_{i=1}^{n}q(w_i)Conditions: none
Rule: L_product_form · Uses: G1, G2
Leaf basis: mul_comm + mul_assoc + power_laws
By Lemma L_product_form (proved in S02.1), the cylinder probability is the product of the position probabilities q(w_i). This product form is the key to the consistency and marginal computations that follow.
S03 · equivalent
Before
c(w0)+c(w1)
After
c(w)
Conditions: none
Rule: L_consistency · Uses: S02
Leaf basis: distributive + add_comm + mul_one
By Lemma L_consistency (proved in S03.1), summing words of length n+1 over the last position gives back the probability at length n. This is exactly the consistency required by the Kolmogorov existence theorem.
S04 · induction_step
Before
\sum_{w\in\{0,1\}^n}c(w)After
1\quad\text{for every}\ n\ge1Conditions: none
Rule: nat_induction · Uses: S03, G1
Leaf basis: nat_induction + finite_sum_rules + numeric_evaluation
Induct on n. Base n=1: c(0)+c(1)=2/3+1/3=1 (N8). Step: the words of length n+1 are exactly the words w of length n followed by 0 or 1, so Σ_{|v|=n+1} c(v)=Σ_{|w|=n}(c(w0)+c(w1))=Σ_{|w|=n} c(w)=1 (S03 and the induction hypothesis). Example n=3: the 8 words sum to 8/27+12/27+6/27+1/27=1 (N2, N10). So every level is a probability distribution.
S05 · theorem_application
Before
(\mu_n)_{n\ge1}\ \text{consistent, each a probability}After
\exists!\ P\ \text{on}\ \{0,1\}^{\mathbb{N}}\ \text{with cylinder probabilities}\ c(w)Conditions: \text{consistency (S03)}; \text{total mass 1 (S04)}
Rule: kolmogorov_ext · Uses: S03, S04
Leaf basis: kolmogorov_ext
Let μ_n(w)=c(w). S04 shows each μ_n is a probability on {0,1}^n, and S03 gives consistency. By the cited Kolmogorov existence theorem there is a unique probability on {0,1}^ℕ with cylinder probabilities c(w); X_i is the i-th coordinate. This guide checks only the theorem's hypotheses and does not prove the theorem itself.
S06 · equivalent
Before
P(X_j=a_j\ \forall j\in J),\ n=\max J
After
\sum_{w\in\{0,1\}^n,\ w_j=a_j\ (j\in J)}c(w)Conditions: J\ \text{finite, nonempty}
Rule: prob_axioms · Uses: S05, G2, prob_axioms
Leaf basis: prob_axioms + set_operations_def + finite_sum_rules
Take n=max J. The event 'X_j=a_j for all j∈J' is the union of the cylinders of the words of length n that take the prescribed values on J, and different words give disjoint cylinders. By finite additivity of probability, its probability is the sum of these c(w).
S07 · equivalent
Before
\sum_{w_J=a}\prod_{i=1}^{n}q(w_i)After
\prod_{j\in J}q(a_j)Conditions: none
Rule: L_marginal · Uses: S06, S02
Leaf basis: distributive + finite_sum_rules + nat_induction
Rewrite each c(w) in the sum by the product form (S02), then apply Lemma L_marginal (proved in S07.1 and S07.2): summing over each free position produces a factor q(0)+q(1)=1, and only the factors of the prescribed positions remain.
S08 · implies
Before
J=\{i\}After
P(X_i=1)=\tfrac13,\ P(X_i=0)=\tfrac23\quad\text{for every}\ iConditions: none
Rule: forall_intro · Uses: S06, S07, G1
Leaf basis: forall_intro + forall_elim
Take J={i}: P(X_i=a)=q(a), so P(X_i=1)=1/3 and P(X_i=0)=2/3, the same for every i; 1/3+2/3=1 (N8). This is the complete marginal law (the probabilities of both values), not just the mean E[X_i]=1/3. So (a), identical distribution, holds.
S09 · implies
Before
J\ \text{finite},\ (a_j)_{j\in J}After
P(X_j=a_j\ \forall j\in J)=\prod_{j\in J}P(X_j=a_j)Conditions: none
Rule: forall_intro · Uses: S06, S07, S08, G3
Leaf basis: forall_intro + eq_replace
First split cases. If J is empty, the event with no coordinate constraints is all of Ω and has probability 1; the right side is also 1 by the empty-product convention. The independence equality holds without taking max of the empty set. If J is finite and nonempty, S06 takes n=max J. Then S06 and S07 give Π_{j∈J} q(a_j) on the left, and S08 identifies each q(a_j) with P(X_j=a_j), giving the product of marginals. These cases cover every finite J, so (b) holds for every finite subfamily, not just pairs.
S10 · implies
Before
P(X_2=1,\ X_5=0)
After
\tfrac13\cdot\tfrac23=\tfrac29
Conditions: none
Rule: forall_elim · Uses: S09, S08
Leaf basis: forall_elim + numeric_evaluation
Instance: J={2,5}, a_2=1, a_5=0. S09 gives (1/3)(2/3)=2/9. Direct check from S06: n=5, the free positions 1, 3, 4 give 2^3=8 words, summing to (1/3)(2/3)((2/3)+(1/3))^3=2/9 (N4).
S11 · implies
Before
Z_1,Z_2\ \text{IID Bernoulli}(\tfrac13),\ Z_i=Z_1\ (i\ge3)After
P(Z_1=1,Z_3=1)=\tfrac13\neq\tfrac19
Conditions: none
Rule: quantifier_negation · Uses: G3
Leaf basis: quantifier_negation + numeric_evaluation
Contrast (prefix): let Z_1, Z_2 be independent Bernoulli(1/3) and let every later Z_i equal Z_1. The length-2 prefix satisfies the product formula and every Z_i has marginal Bernoulli(1/3); but P(Z_1=1,Z_3=1)=P(Z_1=1)=1/3, while P(Z_1=1)P(Z_3=1)=1/9 (N5), so independence fails for J={1,3}. A checked finite prefix cannot prove that the infinite sequence is IID.
S12 · implies
Before
U_1,U_2\ \text{IID Bernoulli}(\tfrac12),\ U_3=U_1\oplus U_2After
\text{pairwise independent, but}\ P(U_1=U_2=U_3=1)=0\neq\tfrac18Conditions: none
Rule: quantifier_negation · Uses: G3
Leaf basis: quantifier_negation + prob_axioms + numeric_evaluation
Contrast (pairwise): U_3 is the sum of U_1 and U_2 modulo 2. The four values of (U_1,U_2) each have probability 1/4. For any a, b, P(U_1=a,U_3=b)=P(U_1=a,U_2=a⊕b)=1/4=(1/2)(1/2) (N7); likewise for (U_2,U_3), and all three marginals are Bernoulli(1/2). But U_1=U_2=1 forces U_3=0, so P(U_1=U_2=U_3=1)=0≠1/8 (N6). Pairwise independence plus identical distribution is still not IID.
S13 · implies
Before
V\equiv\tfrac13
After
E[V]=E[X_1]=\tfrac13,\ \text{but}\ P(V=1)=0\neq\tfrac13Conditions: none
Rule: quantifier_negation · Uses: S08, G3
Leaf basis: quantifier_negation + numeric_evaluation
Contrast (equal expectations): the constant V≡1/3 has the same mean 1/3 as X_1, but P(V=1)=0 while P(X_1=1)=1/3. So equal expectations do not give equal laws; (a) compares complete marginals (S08).
S14 · implies
Before
\text{S05, S08, S09, S11 to S13}After
(X_i)_{i\ge1}\ \text{IID Bernoulli}(\tfrac13)Conditions: none
Rule: mko_iid_def · Uses: S05, S08, S09, S11, S12, S13
Leaf basis: and_intro_elim + def_unfold
Under the probability of S05, every X_i has complete marginal Bernoulli(1/3) (S08) and every finite subfamily is independent (S09), so (X_i) is IID; this rests on the symbolic certificate valid for all n and all words, not on any prefix. The three contrasts (S11 to S13) show that the weaker checks are insufficient.
Substeps and lemmas
S02.1 · equivalent
Before
\prod_{i=1}^{n}q(w_i)After
p^{k(w)}(1-p)^{n-k(w)}Conditions: none
Rule: power_laws · Uses: G1, mul_comm, power_laws
Leaf basis: mul_comm + mul_assoc + power_laws
In the product, the k(w) positions with w_i=1 each contribute q(1)=p, and the n−k(w) positions with w_i=0 each contribute q(0)=1−p. Commutativity and associativity gather equal factors, giving p^{k(w)}(1−p)^{n−k(w)}. S02.1 proves Lemma L_product_form.
S03.1 · equivalent
Before
c(w)q(0)+c(w)q(1)
After
c(w)\big((1-p)+p\big)=c(w)
Conditions: none
Rule: distributive · Uses: S02, distributive
Leaf basis: distributive + add_inverse + add_comm + mul_one
By the product form (S02), c(wa)=c(w)q(a). So c(w0)+c(w1)=c(w)(1−p)+c(w)p=c(w)((1−p)+p)=c(w)·1=c(w) (N3 checks this as a polynomial identity). S03.1 proves Lemma L_consistency.
S07.1 · induction_base
Before
J=\{1,2,\text{up to }n\}After
\sum=\prod_{i=1}^{n}q(a_i)Conditions: none
Rule: nat_induction · Uses: S02
Leaf basis: nat_induction + finite_sum_rules
Induct on the number f=n−|J| of free positions. Base f=0: every position is prescribed, the sum has the single word w=a, and its value is Π_{i=1}^{n} q(a_i)=Π_{j∈J} q(a_j).
S07.2 · induction_step
Before
f\ \text{free positions},\ i\ \text{one of them}After
\sum=\big(q(0)+q(1)\big)\cdot\big(\text{sum with}\ f-1\ \text{free positions}\big)=\prod_{j\in J}q(a_j)Conditions: none
Rule: nat_induction · Uses: S07.1, S02, distributive
Leaf basis: nat_induction + distributive + finite_sum_rules + add_inverse
Step: let f≥1 and pick a free position i. Split the sum by the value of w_i; each term has the factor q(w_i), which distributivity pulls out: sum=Σ_{b∈{0,1}} q(b)·T=(q(0)+q(1))T=T, where T is the sum, under the same constraints, of the products over the other positions, with only f−1 free positions. By the induction hypothesis T=Π_{j∈J} q(a_j). S07.1 and S07.2 together prove Lemma L_marginal.