Main derivation
S01 · definition
Before
z\in\mathbb{Z}After
z=0\ \lor\ z\in N^{+}\ \lor\ \exists n\in N^{+}:\ z=-nConditions: none
Rule: mko_int_def · Uses: G1, G2, mko_int_def
Leaf basis: mko_int_def + def_unfold + set_operations_def
By the MKO, Z consists of three parts: 0, the positive natural numbers, and the negatives of positive natural numbers. So deciding z∈Z means deciding whether at least one of three disjuncts holds. Here −n is the additive inverse of n, not the action 'subtract n' (G3).
S02 · implies
Before
7=0+1+1+1+1+1+1+1
After
7\in N^{+},\ -7\in\mathbb{Z}Conditions: none
Rule: exists_intro · Uses: S01, G1
Leaf basis: exists_intro + nat_least_inductive + set_operations_def + numeric_evaluation
7 is obtained from 0 by seven successor steps (N1), so applying the closure of G1 seven times gives 7∈N; 7≠0, so 7∈N⁺. Taking n=7, −7=−n, so the third disjunct holds and by ∃-introduction −7∈Z.
S03 · implies
Before
\text{decimal numeral }3.0After
3\cdot10^0+0\cdot10^{-1}=3\in N^{+}\subseteq\mathbb{Z}Conditions: none
Rule: positional_def · Uses: S01, G4, positional_def
Leaf basis: positional_def + power_laws + numeric_evaluation + set_operations_def
The decimal numeral '3.0' denotes 3·10⁰+0·10⁻¹=3. 3 is obtained from 0 by three successor steps, so 3∈N and 3≠0, hence 3∈N⁺; the second disjunct holds and 3∈Z. The decimal point in the numeral does not change the value; integrality is decided by the number, not by how the numeral looks.
S04 · equivalent
Before
-2.5\in\mathbb{Z}After
-2.5=0\ \lor\ -2.5\in N^{+}\ \lor\ \exists n\in N^{+}:\ -2.5=-nConditions: none
Rule: def_unfold · Uses: S01, G4
Leaf basis: def_unfold + mko_int_def
Unfold the membership condition of S01 for −2.5 (that is −5/2, N3). To prove −2.5∉Z, all three disjuncts must be excluded.
S05 · equivalent
Before
-2.5=-n
After
n=2.5
Conditions: none
Rule: eq_congruence · Uses: G3, eq_congruence
Leaf basis: eq_congruence + add_inverse + add_comm + add_assoc + add_zero
Add n+2.5 to both sides: the left side −2.5+(n+2.5)=n and the right side −n+(n+2.5)=2.5 (commutativity, associativity and inverses for addition), so n=2.5. Backward: if n=2.5, taking additive inverses of both sides gives −n=−2.5. So '−2.5 equals some −n' is exactly 'n=2.5'.
S06 · implies
Before
T_3=\{0,1,2\}\cup\{x\in\mathbb{R}\mid x\ge 3\}After
N\subseteq T_3
Conditions: none
Rule: L_N_T3 · Uses: G1, L_N_T3
Leaf basis: L_N_T3
By Lemma L_N_T3, every natural number is 0, 1 or 2, or else ≥3. So there is no natural number strictly between 2 and 3, and every natural number is ≥0. The lemma is proved in S06.1 and S06.2.
S07 · implies
Before
N\subseteq T_3
After
2.5\notin N,\ -2.5\notin N
Conditions: none
Rule: contrapositive · Uses: S06, G4
Leaf basis: contrapositive + subset_def + order_basics + numeric_evaluation
2.5 differs from 0, 1 and 2, and 2.5<3, so 2.5∉T₃; −2.5<0 while every element of T₃ is ≥0, so −2.5∉T₃. Since N⊆T₃, the contrapositive gives 2.5∉N and −2.5∉N.
S08 · implies
Before
\text{the three disjuncts of S04}After
-2.5\notin\mathbb{Z}Conditions: none
Rule: de_morgan_logic · Uses: S04, S05, S07
Leaf basis: de_morgan_logic + set_operations_def + quantifier_negation
First disjunct: −2.5≠0. Second: −2.5∉N, so it is not in N⁺⊆N either. Third: by S05 it amounts to '2.5∈N⁺', but 2.5∉N (S07). All three fail, so by De Morgan's law the disjunction is false, and by the equivalence S04, −2.5∉Z.
S09 · implies
Before
(-a)+a=0,\ (-a)+(-(-a))=0
After
-(-a)=a\quad(a\in\mathbb{R})Conditions: none
Rule: L_inv_unique · Uses: G3, L_inv_unique
Leaf basis: L_inv_unique + add_comm + add_inverse + forall_intro
For any real a: by commutativity (−a)+a=a+(−a)=0, so a is an additive inverse of −a; by definition −(−a) is also an additive inverse of −a. By Lemma L_inv_unique (with the lemma's a taken to be −a), −(−a)=a. a was arbitrary, so this holds for every real. The lemma is proved in S09.1 and S09.2.
S10 · implies
Before
-(-4)
After
4\in N^{+}\subseteq\mathbb{Z}Conditions: none
Rule: forall_elim · Uses: S09, S01
Leaf basis: forall_elim + numeric_evaluation + set_operations_def
Take a=4 in S09: −(−4)=4. Check: (−4)+4=0 (N8). 4 is obtained from 0 by four successor steps and 4≠0, so 4∈N⁺⊆Z; −4 itself is also an integer (third part, n=4).
S11 · implies
Before
z\in\mathbb{Z}After
-z\in\mathbb{Z}Conditions: none
Rule: case_analysis · Uses: S01, S09
Leaf basis: case_analysis + L_inv_unique + forall_intro + exists_intro
Split into the three cases of S01. (i) z=0: 0+0=0, so 0 is an inverse of 0 and by uniqueness −0=0∈Z. (ii) z∈N⁺: with n=z, −z lies in the third part. (iii) z=−n with n∈N⁺: by S09, −z=−(−n)=n∈N⁺⊆Z. The three cases cover every element of Z, so −z∈Z for every integer z. This is a proof for an arbitrary z, not a generalization from a few examples.
S12 · implies
Before
5-8=5+(-8)
After
5+(-8)=-3
Conditions: none
Rule: L_inv_unique · Uses: G3, L_inv_unique
Leaf basis: L_inv_unique + add_comm + add_assoc + add_inverse + numeric_evaluation
By G3, binary subtraction 5−8 is defined as 5+(−8). Compute 3+(5+(−8))=(3+5)+(−8)=8+(−8)=0 (associativity and inverses), and also 3+(−3)=0; by Lemma L_inv_unique (a=3), 5+(−8)=−3. Two kinds of '−' appear: in 5−8 it is a binary operation (subtraction), in −3 a unary sign (the inverse of 3); the result of the operation is the number −3, and −3∈Z (3∈N⁺).
S13 · implies
Before
\text{S02, S03, S08, S10, S11, S12}After
-7,\ 3,\ 4,\ -3\in\mathbb{Z};\ -2.5\notin\mathbb{Z};\ \forall z\in\mathbb{Z}:\ -z\in\mathbb{Z}Conditions: none
Rule: and_intro_elim · Uses: S02, S03, S08, S10, S11, S12
Leaf basis: and_intro_elim
−7, 3 (written 3.0), −(−4)=4 and 5−8=−3 are integers; −2.5 is not; and Z is closed under negation.
Substeps and lemmas
S06.1 · implies
Before
0\in T_3;\ x\in\{0,1,2\}After
x+1\in\{1,2,3\}\subseteq T_3Conditions: none
Rule: case_analysis · Uses: G1, nat_least_inductive
Leaf basis: case_analysis + numeric_evaluation + order_basics
0∈T₃. For x=0, 1, 2, x+1 is 1, 2, 3 respectively; 1 and 2 are in {0,1,2} and 3≥3, so all lie in T₃.
S06.2 · implies
Before
x\ge 3
After
x+1\ge 4\ge 3;\ T_3\ \text{inductive},\ N\subseteq T_3Conditions: none
Rule: nat_least_inductive · Uses: S06.1, nat_least_inductive
Leaf basis: nat_least_inductive + case_analysis + order_basics
If x≥3 then x+1≥4≥3, so x+1∈T₃. Every element of T₃ is in {0,1,2} or is ≥3, and the two cases (S06.1 and here) cover them all, so T₃ is closed under x↦x+1; with 0∈T₃, T₃ is inductive. By the minimality in G1, N⊆T₃. S06.1 and S06.2 together prove Lemma L_N_T3.
S09.1 · implies
Before
a+x=0,\ a+y=0
After
x=x+0=x+(a+y)=(x+a)+y
Conditions: none
Rule: add_assoc · Uses: add_zero, add_assoc
Leaf basis: add_zero + eq_replace + add_assoc
x=x+0 (additive identity) =x+(a+y) (replace 0 using the hypothesis a+y=0) =(x+a)+y (associativity).
S09.2 · implies
Before
(x+a)+y
After
y,\ \text{hence}\ x=yConditions: none
Rule: add_comm · Uses: S09.1, add_comm
Leaf basis: add_comm + eq_replace + add_zero + eq_refl_symm_trans
(x+a)+y=(a+x)+y (commutativity) =0+y (replace using the hypothesis a+x=0) =y. Chaining with S09.1 gives x=y. S09.1 and S09.2 together prove Lemma L_inv_unique.