Main derivation
S01 · definition
Before
(a_n)
After
a:\mathbb{N}\to\mathbb{R},\ a_0=5,\ a_{n+1}=a_n+3Conditions: 0\in\mathbb{N}
Rule: mko_seq_def · Uses: G1, G2, mko_seq_def
Leaf basis: mko_seq_def + function_def
By the MKO a sequence is a function with domain ℕ: the index n is the input and a_n the output. The given information is a recurrence with an initial condition; it constrains the values of a but does not yet say that such an a exists or is unique, which is what S02 to S06 prove.
S02 · definition
Before
n\in\mathbb{N}After
s:\mathbb{N}\to\mathbb{R},\ s_n=5+3nConditions: none
Rule: def_intro · Uses: S01
Leaf basis: def_intro + function_def
Propose the candidate sequence s_n=5+3n. For each natural number n, 5+3n is a single well-defined real number, so s is a function ℕ→ℝ, that is, a sequence.
S03 · equivalent
Before
s_0
After
s_0=5+3\cdot 0=5
Conditions: none
Rule: eq_replace · Uses: S02
Leaf basis: eq_replace + mul_zero + add_zero
Substitute n=0: 3·0=0 (multiplication by zero) and 5+0=5 (additive identity), so s₀=5, as the initial condition requires.
S04 · equivalent
Before
s_{n+1}=5+3(n+1)After
s_{n+1}=(5+3n)+3=s_n+3Conditions: none
Rule: distributive · Uses: S02
Leaf basis: distributive + add_assoc + mul_one
For any n: 5+3(n+1)=5+(3n+3) by distributivity, =(5+3n)+3 by associativity, =s_n+3. So s satisfies the recurrence, as an identity valid for every n.
S05 · implies
Before
s_0=5,\ s_{n+1}=s_n+3After
\exists\ a:\mathbb{N}\to\mathbb{R}\ \text{with}\ a_0=5,\ a_{n+1}=a_n+3Conditions: none
Rule: exists_intro · Uses: S02, S03, S04
Leaf basis: exists_intro + and_intro_elim
s is a sequence (S02) satisfying the initial condition (S03) and the recurrence (S04), so at least one sequence satisfies G2. Existence is obtained by exhibiting an example.
S06 · theorem_application
Before
a_0=5,\ a_{n+1}=a_n+3After
a_n=5+3n\ \text{for every}\ n\in\mathbb{N},\ \text{so}\ a=sConditions: none
Rule: L_unique · Uses: S01, S02, L_unique, function_def
Leaf basis: L_unique + function_def + nat_induction
By Lemma L_unique, proved by induction in S06.1 and S06.2, every sequence satisfying G2 agrees with s at every index; both have domain ℕ and codomain ℝ, so a=s by function_def. Together with S05, exactly one such sequence exists.
S07 · implies
Before
a_n=5+3n
After
a_0=5,\ a_1=8,\ a_2=11,\ a_3=14,\ a_4=17,\ a_{10}=35Conditions: none
Rule: forall_elim · Uses: S06
Leaf basis: forall_elim + numeric_evaluation
Substitute n=0, 1, 2, 3, 4 and 10 into the general term: 5, 8, 11, 14, 17 and 35. The initial term is a₀=5 (index 0), not a₁.
S08 · implies
Before
a_{10}=35After
a_{10}\ \text{is the 11th value};\ b_k=a_{k-1}\ (k\geq 1)\ \text{gives}\ b_{10}=a_9=32Conditions: none
Rule: def_unfold · Uses: S07, G1
Leaf basis: def_unfold + numeric_evaluation
The indices 0 to 10 number 11, so a₁₀ is the 11th value when counting from 1. If someone uses the 1-based labels b_k=a_{k−1} (k≥1, so b₁=a₀=5), their 'tenth term' is b₁₀=a₉=5+27=32, not a₁₀=35. Speaking of 'the tenth term' requires saying whether indexing starts at 0 or at 1.
S09 · definition
Before
L=4
After
(a_0,a_1,a_2,a_3)=(5,8,11,14),\ \ \text{indices}\ \{0,1,2,3\}Conditions: 0\leq n<4
Rule: mko_seq_def · Uses: G3, S07, mko_seq_def
Leaf basis: mko_seq_def + set_builder
The finite sequence of length L=4 has index set {n∈ℕ ∣ n<4}={0,1,2,3}; it is a function {0,1,2,3}→ℝ with values 5, 8, 11, 14. The index 4 is not permitted (4<4 fails), so this finite sequence has no a₄, even though the infinite sequence has a₄=17.
S10 · implies
Before
c_n=5+3n+n(n-1)(n-2)
After
c_0=5,\ c_1=8,\ c_2=11
Conditions: none
Rule: mul_zero · Uses: S07
Leaf basis: mul_zero + add_zero + numeric_evaluation
Consider another sequence c_n=5+3n+n(n−1)(n−2). For n=0, 1, 2 one factor of the product n(n−1)(n−2) is 0, so the product is 0 and c_n=5+3n. Hence c has exactly the same prefix 5, 8, 11 as a.
S11 · implies
Before
c_3
After
c_3=5+9+3\cdot 2\cdot 1=20\neq 14=a_3,\ \text{so}\ c\neq aConditions: none
Rule: function_def · Uses: S07, S10, function_def
Leaf basis: function_def + numeric_evaluation
c₃=5+3·3+3·2·1=14+6=20, while a₃=14. The two sequences differ at index 3, so c≠a by function_def. Hence the prefix 5, 8, 11 does not determine the sequence: at least two different sequences begin with it.
S12 · implies
Before
c_3-c_2
After
c_3-c_2=20-11=9\neq 3
Conditions: none
Rule: contrapositive · Uses: S06, S10, S11
Leaf basis: contrapositive + numeric_evaluation
c satisfies c₀=5 but fails the recurrence at n=2: c₃−c₂=9≠3. This agrees with S06: only a satisfies both the initial condition and the recurrence. What determines the sequence is the rule together with the initial condition, not finitely many first terms.
S13 · implies
Before
S05\ \text{to}\ S09,\ S11After
a_n=5+3n\ \text{unique};\ a_{10}=35;\ (5,8,11,14)\ \text{on}\ \{0,1,2,3\};\ \text{prefix not determining}Conditions: none
Rule: and_intro_elim · Uses: S05, S06, S07, S08, S09, S11
Leaf basis: and_intro_elim
Goal (1) is completed by S05 and S06, (2) by S07 and S08, (3) by S09, and (4) by S10 and S11.
Error controls and negative controls
State whether indexing starts at 0 or at 1
The first mistake named by the MKO is confusing indexing from 0 with indexing from 1. Here a₀=5 is the initial term and a₁₀=35; with 1-based labels b_k=a_{k−1}, the 'tenth term' is b₁₀=32. The same list of numbers gives different 'tenth terms' under the two labelings, so the convention must be stated first.
A finite prefix does not determine a sequence
The second mistake named by the MKO is claiming that a prefix pins down a unique sequence. After 5, 8, 11, a continues with 14 and c with 20; in fact any finite prefix can be followed by any values. What determines a sequence is an explicit rule (a general term, or a recurrence with initial conditions), and S06 proves uniqueness from exactly such a rule.
A finite sequence has indices only up to L−1
The finite sequence of length L=4 has the 4 indices 0, 1, 2, 3. Writing 1 to 4 loses a₀ and adds a non-existent a₄; writing 0 to 4 gives 5 indices, so length 5. a₄=17 belongs to the infinite sequence, not to this finite one.
- From 5, 8, 11 the next term must be 14 →
reject — c_n=5+3n+n(n−1)(n−2) has the same prefix but c₃=20. - Under 1-based labels the tenth term is 35 →
reject — b₁₀=a₉=32; 35 is a₁₀, the 11th value. - The finite sequence (5,8,11,14) of length 4 has a term a₄=17 →
reject — Its indices are only 0, 1, 2, 3; 4<4 fails. - The finite sequences (5,8) and (8,5) are equal because both have value set {5,8} →
reject — A sequence is a function; the values at index 0 are 5≠8. The order of indices is part of the data. - c also satisfies c₀=5 and c_{n+1}=c_n+3 →
reject — c₃−c₂=9≠3.