Main derivation
S01 · definition
Before
a,b\in\mathbb{R}After
a<b\ :\iff\ a\le b\land a\neq b;\ \text{exactly one of}\ a<b,\ a=b,\ b<aConditions: none
Rule: mko_order · Uses: G1, mko_order
Leaf basis: mko_order + order_basics
Define the strict order as in the MKO. Totality (a≤b or b≤a) and antisymmetry (both together force a=b) give trichotomy: exactly one of a<b, a=b, b<a holds. This is the MKO's three-way comparison with outputs <, = or >.
S02 · equivalent
Before
a<b
After
a+c<b+c
Conditions: none
Rule: L_translate · Uses: S01
Leaf basis: order_basics + add_assoc + add_inverse + add_zero
By Lemma L_translate, adding c to both numbers does not change the comparison. ⇒ is S02.1 and ⇐ is S02.2.
S03 · equivalent
Before
\frac{p}{q}<\frac{r}{s}After
ps<rq
Conditions: q>0; s>0
Rule: L_cross · Uses: S01
Leaf basis: order_basics + frac_rules + mul_inverse_nonzero
By Lemma L_cross: when both denominators are positive, comparing two fractions is the same as comparing the cross products. The proof is in S03.1 to S03.3; the positive-denominator condition cannot be dropped.
S04 · equivalent
Before
-\tfrac{7}{12}\ ?\ -\tfrac{5}{9}After
(-7)\cdot9=-63<-60=(-5)\cdot12\ \Rightarrow\ -\tfrac{7}{12}<-\tfrac{5}{9}Conditions: 12>0; 9>0
Rule: L_cross · Uses: S03, G2
Leaf basis: order_basics + numeric_evaluation
Take p=−7, q=12, r=−5, s=9; both denominators are positive. ps=−63 and rq=−60, and −60−(−63)=3>0, so −63<−60; by L_cross, −7/12<−5/9.
S05 · equivalent
Before
-\tfrac{5}{9}\ ?\ -\tfrac{1}{2}After
(-5)\cdot2=-10<-9=(-1)\cdot9\ \Rightarrow\ -\tfrac{5}{9}<-\tfrac{1}{2}Conditions: 9>0; 2>0
Rule: L_cross · Uses: S03, G2
Leaf basis: order_basics + numeric_evaluation
Take p=−5, q=9, r=−1, s=2: ps=−10, rq=−9 and −10<−9, so −5/9<−1/2.
S06 · equivalent
Before
-\tfrac{1}{2}\ ?\ \tfrac{1}{3}After
(-1)\cdot3=-3<2=1\cdot2\ \Rightarrow\ -\tfrac{1}{2}<\tfrac{1}{3}Conditions: 2>0; 3>0
Rule: L_cross · Uses: S03, G2
Leaf basis: order_basics + numeric_evaluation
Take p=−1, q=2, r=1, s=3: ps=−3, rq=2 and −3<2, so −1/2<1/3.
S07 · implies
Before
-\tfrac{7}{12}<-\tfrac{5}{9},\ -\tfrac{5}{9}<-\tfrac{1}{2},\ -\tfrac{1}{2}<\tfrac{1}{3}After
-\tfrac{7}{12}<-\tfrac{5}{9}<-\tfrac{1}{2}<\tfrac{1}{3}Conditions: none
Rule: order_basics · Uses: S04, S05, S06
Leaf basis: order_basics + logic_chain
The three comparisons form a chain; by transitivity of <, each number in the chain is less than every number to its right, for example −7/12<−1/2, −7/12<1/3 and −5/9<1/3. All four elements of M are now ordered.
S08 · implies
Before
M\subseteq\mathbb{R}\ \text{finite},\ M\neq\varnothingAfter
M\ \text{has a minimum}Conditions: M\neq\varnothing
Rule: L_finite_min · Uses: S01
Leaf basis: nat_induction + order_basics + case_analysis
The MKO asserts that a nonempty finite set of reals has a minimum and a maximum. By Lemma L_finite_min, the minimum exists, proved by induction on the number of elements n≥1 (S08.1, S08.2); the argument for the maximum is the same with ≥ in place of ≤.
S09 · implies
Before
-\tfrac{7}{12}\in M,\ -\tfrac{7}{12}\le x\ \forall x\in MAfter
\min M=-\tfrac{7}{12}Conditions: none
Rule: min_max_def · Uses: S07, G2, min_max_def
Leaf basis: min_max_def + order_basics
−7/12 belongs to M (listed third), by S07 it is less than the other three elements, and −7/12≤−7/12 for itself, so it is the minimum. M lists −5/9 first, but extrema are decided by value, not by the order of listing.
S10 · implies
Before
\tfrac{1}{3}\in M,\ \tfrac{1}{3}\ge x\ \forall x\in MAfter
\max M=\tfrac{1}{3}Conditions: none
Rule: min_max_def · Uses: S07, G2, min_max_def
Leaf basis: min_max_def + order_basics
1/3 belongs to M and by S07 the other three elements are all less than 1/3, so 1/3 is the maximum.
S11 · implies
Before
m,m'\ \text{both minima of}\ MAfter
m=m'
Conditions: none
Rule: L_min_unique · Uses: S09, S10
Leaf basis: min_max_def + order_basics
By Lemma L_min_unique the minimum is unique, so one may speak of 'the' minimum −7/12; the maximum 1/3 is likewise unique. The proof is in S11.1.
S12 · implies
Before
l=-\tfrac{7}{12},\ u=\tfrac{1}{3}After
l\le u
Conditions: none
Rule: order_basics · Uses: S07, S09, S10, G3
Leaf basis: order_basics + mko_order
MKO assumption a3 asks first to check that the lower endpoint of a closed interval is no greater than the upper one. By S07, l=−7/12<1/3=u, so l≤u and [l,u] is a valid closed interval.
S13 · implies
Before
l\le l\le u,\ l\le u\le u
After
l\in[l,u],\ u\in[l,u],\ M\subseteq[l,u]
Conditions: l\le u
Rule: interval_def · Uses: S12, S09, S10, interval_def
Leaf basis: interval_def + mko_order
Membership in a closed interval is l≤x≤u. For x=l: l≤l (reflexivity) and l≤u (S12), so the left endpoint belongs to the interval; the right endpoint likewise. Every element x of M satisfies min M≤x≤max M (S09, S10), so M⊆[l,u].
S14 · implies
Before
\tfrac{1}{3}<\tfrac{1}{2}\ \ (1\cdot2<1\cdot3)After
\tfrac12\notin[l,u]
Conditions: none
Rule: interval_def · Uses: S03, S13, interval_def
Leaf basis: interval_def + contradiction + order_basics
By L_cross (p=1, q=3, r=1, s=2): 1·2=2<3=1·3, so 1/3<1/2. If 1/2≤1/3, antisymmetry together with 1/3≤1/2 would give 1/2=1/3, contradicting 1/3≠1/2; so 1/2≤u fails and 1/2 is not in the interval.
S15 · equivalent
Before
-\tfrac{7}{12}<-\tfrac{5}{9}After
-\tfrac{7}{12}+\tfrac{7}{12}<-\tfrac{5}{9}+\tfrac{7}{12},\ \text{i.e.}\ 0<\tfrac{1}{36}Conditions: none
Rule: L_translate · Uses: S02, S04
Leaf basis: order_basics + frac_rules + numeric_evaluation
Apply L_translate with c=7/12: −7/12<−5/9 ⇔ 0<−5/9+7/12. Compute −5/9+7/12=−20/36+21/36=1/36>0, consistent with S04. Translation does not change the comparison, so this is a rewriting, not a new kind of exercise.
S16 · implies
Before
\text{S02 to S15}After
\min M=-\tfrac{7}{12},\ \max M=\tfrac13,\ l,u\in[l,u],\ \tfrac12\notin[l,u]Conditions: none
Rule: and_intro_elim · Uses: S02, S03, S08, S09, S10, S11, S13, S14
Leaf basis: and_intro_elim
Summary: translation invariance and positive-denominator cross-comparison are proved; a nonempty finite set has a unique minimum; M is ordered −7/12<−5/9<−1/2<1/3 with minimum −7/12 and maximum 1/3; the closed interval [−7/12,1/3] contains both endpoints and all of M, but not 1/2.
Substeps and lemmas
S02.1 · implies
Before
a<b
After
a+c<b+c
Conditions: none
Rule: order_basics · Uses: order_basics
Leaf basis: order_basics
(⇒) This is exactly the order axiom 'u<v ⇒ u+w<v+w' with u=a, v=b, w=c.
S02.2 · implies
Before
a+c<b+c
After
(a+c)+(-c)<(b+c)+(-c),\ \text{i.e.}\ a<bConditions: none
Rule: order_basics · Uses: order_basics, add_inverse
Leaf basis: order_basics + add_assoc + add_inverse + add_zero
(⇐) Apply the same axiom to a+c<b+c with w=−c to get (a+c)+(−c)<(b+c)+(−c); simplifying both sides (associativity, c+(−c)=0, adding 0) gives a<b. S02.1 and S02.2 together prove Lemma L_translate.
S03.1 · implies
Before
q>0,\ s>0
After
qs>0,\ \tfrac{1}{qs}>0Conditions: q>0; s>0
Rule: order_basics · Uses: order_basics, mul_inverse_nonzero
Leaf basis: order_basics + mul_zero + mul_inverse_nonzero + contradiction
Multiply 0<q by s>0 to get 0·s<q·s, i.e. qs>0. Then qs≠0, so 1/(qs) exists; if 1/(qs)<0, multiplying by qs>0 would give 1<0, a contradiction, and 1/(qs)≠0 since (1/(qs))·qs=1, so 1/(qs)>0.
S03.2 · implies
Before
\frac{p}{q}<\frac{r}{s}After
\frac{p}{q}\cdot qs<\frac{r}{s}\cdot qs,\ \text{i.e.}\ ps<rqConditions: qs>0
Rule: order_basics · Uses: S03.1, frac_rules
Leaf basis: order_basics + frac_rules + mul_comm + mul_assoc
(⇒) Multiply both sides by the positive number qs, which keeps the order; simplify (p/q)·qs=ps and (r/s)·qs=rq (fraction rules, q, s≠0).
S03.3 · implies
Before
ps<rq
After
\frac{ps}{qs}<\frac{rq}{qs},\ \text{i.e.}\ \frac{p}{q}<\frac{r}{s}Conditions: \tfrac{1}{qs}>0
Rule: order_basics · Uses: S03.1, frac_rules
Leaf basis: order_basics + frac_rules
(⇐) Multiply both sides by the positive number 1/(qs) (S03.1), which keeps the order; cancel ps/(qs)=p/q and rq/(qs)=r/s. S03.1 to S03.3 together prove Lemma L_cross.
S08.1 · induction_base
Before
M=\{a\}After
\min M=a
Conditions: none
Rule: nat_induction · Uses: G3, min_max_def
Leaf basis: min_max_def + order_basics
Base n=1: M={a}. a∈M, and the only element x=a satisfies a≤a (reflexivity), so a is the minimum (and also the maximum).
S08.2 · induction_step
Before
|M|=n+1,\ a\in M,\ m'=\min(M\setminus\{a\})After
\min M=m'\ \text{if}\ m'\le a,\ \ \min M=a\ \text{if}\ a<m'Conditions: n\ge1
Rule: nat_induction · Uses: S08.1, min_max_def, case_analysis
Leaf basis: nat_induction + case_analysis + order_basics + min_max_def
Inductive step: assume every nonempty set with n elements has a minimum (n≥1). Take |M|=n+1 and a∈M; M′=M∖{a} has n elements and, by the induction hypothesis, a minimum m′. By trichotomy there are two cases: if m′≤a, then m′ is no greater than every element of M′ and than a, so m′ is the minimum of M; if a<m′, then for x∈M′ we have a<m′≤x, so a≤x by transitivity, and a≤a, so a is the minimum. S08.1 and S08.2 together prove Lemma L_finite_min.
S11.1 · implies
Before
m,m'\in M,\ m\le x,\ m'\le x\ \forall x\in M
After
m\le m',\ m'\le m\ \Rightarrow\ m=m'
Conditions: none
Rule: mko_order · Uses: min_max_def, G1
Leaf basis: min_max_def + mko_order
m is a minimum and m′∈M, so m≤m′; likewise m′≤m. By antisymmetry m=m′. The maximum is handled the same way with ≥. This step proves Lemma L_min_unique.