Main derivation
S01 · definition
Before
\text{points and lines of the chosen model}After
\text{point}\in\mathbb{R}^2,\ L(P,Q)=\{P+t(Q-P)\mid t\in\mathbb{R}\}\ (P\neq Q)Conditions: none
Rule: mko_model_dependence · Uses: G1, G2
Leaf basis: mko_model_dependence
The MKO says the precise properties of lines are fixed by the chosen model, so the model is fixed first: points are ordered pairs in ℝ², and lines are given by the parametric form of G2. From here on, 'on the line' means membership in a set of this model, not how a sketch looks.
S02 · implies
Before
P=(1,2),\ Q=(3,6)
After
P\neq Q
Conditions: none
Rule: cartesian_product_def · Uses: G3, cartesian_product_def
Leaf basis: cartesian_product_def + numeric_evaluation
Two ordered pairs are equal only if both coordinates agree; the first coordinates of P and Q are 1≠3, so P≠Q. This is the hypothesis the line definition in G2 requires, so L(P,Q) really is a line.
S03 · equivalent
Before
Q-P
After
(2,4)
Conditions: none
Rule: coord_ops · Uses: G1, G3
Leaf basis: coord_ops + numeric_evaluation
Subtract coordinatewise: Q−P=(3−1, 6−2)=(2,4). Both sides name the same ordered pair, so this is a replacement of equals.
S04 · equivalent
Before
L(P,Q)=\{P+t(Q-P)\mid t\in\mathbb{R}\}After
L=\{(1+2t,\,2+4t)\mid t\in\mathbb{R}\}Conditions: P\neq Q
Rule: coord_ops · Uses: S01, S02, S03, G3
Leaf basis: coord_ops + eq_replace
Substitute P=(1,2) and Q−P=(2,4): P+t(Q−P)=(1,2)+(2t,4t)=(1+2t, 2+4t). For every t the two expressions are the same point, so the two sets are equal. The condition P≠Q was checked in S02.
S05 · equivalent
Before
R\in L
After
\exists t\in\mathbb{R}:\ (4,8)=(1+2t,\,2+4t)Conditions: none
Rule: image_preimage_def · Uses: S04, G3
Leaf basis: image_preimage_def + def_unfold
L is the set of points obtained as t runs through ℝ (the image of t↦(1+2t,2+4t)). So R belongs to L exactly when some real t makes R equal to (1+2t, 2+4t).
S06 · equivalent
Before
(4,8)=(1+2t,\,2+4t)
After
4=1+2t\ \land\ 8=2+4t
Conditions: none
Rule: cartesian_product_def · Uses: S05, cartesian_product_def
Leaf basis: cartesian_product_def
Ordered pairs are equal exactly when both coordinates agree, so the point equation splits into two real equations, and both use the same t.
S07 · equivalent
Before
4=1+2t
After
t=\tfrac{3}{2}Conditions: 2\neq 0
Rule: L_solve_linear · Uses: S06
Leaf basis: eq_congruence + mul_inverse_nonzero + two_nonzero
By Lemma L_solve_linear (a=1, b=2, c=4): 4=1+2t ⇔ t=(4−1)/2=3/2. Dividing by 2 needs 2≠0.
S08 · equivalent
Before
8=2+4t
After
t=\tfrac{3}{2}Conditions: 4\neq 0
Rule: L_solve_linear · Uses: S06
Leaf basis: eq_congruence + mul_inverse_nonzero + numeric_evaluation
The same lemma (a=2, b=4, c=8): 8=2+4t ⇔ t=(8−2)/4=6/4=3/2; here 4≠0.
S09 · implies
Before
4=1+2t\ \land\ 8=2+4t
After
R\in L
Conditions: none
Rule: exists_intro · Uses: S05, S06, S07, S08
Leaf basis: exists_intro + and_intro_elim + logic_chain
Each equation is satisfied exactly by t=3/2, so the single value t=3/2 satisfies both. By the equivalences S05 and S06, R∈L, and indeed R=P+(3/2)(Q−P).
S10 · equivalent
Before
S\in L
After
\exists t\in\mathbb{R}:\ 2=1+2t\ \land\ 5=2+4tConditions: none
Rule: image_preimage_def · Uses: S04, G3
Leaf basis: image_preimage_def + cartesian_product_def
Repeat the two unfoldings of S05 and S06 for S: S∈L exactly when some t satisfies both coordinate equations 2=1+2t and 5=2+4t.
S11 · equivalent
Before
2=1+2t\ \land\ 5=2+4t
After
t=\tfrac{1}{2}\ \land\ t=\tfrac{3}{4}Conditions: 2\neq 0; 4\neq 0
Rule: L_solve_linear · Uses: S10
Leaf basis: eq_congruence + mul_inverse_nonzero + two_nonzero + numeric_evaluation
Solve each equation: 2=1+2t ⇔ t=1/2 and 5=2+4t ⇔ t=3/4. Each equation is transformed by an equivalence, so the conjunction before and after is true or false together.
S12 · implies
Before
\exists t:\ t=\tfrac{1}{2}\ \land\ t=\tfrac{3}{4}After
S\notin L
Conditions: none
Rule: contradiction · Uses: S10, S11
Leaf basis: contradiction + eq_refl_symm_trans + numeric_evaluation
If some t equalled both 1/2 and 3/4, symmetry and transitivity of equality would give 1/2=3/4, but 1/2≠3/4. So no such t exists, and by the equivalence S10, S∉L.
S13 · implies
Before
t=0,\ t=1
After
P\in L,\ Q\in L
Conditions: none
Rule: image_preimage_def · Uses: S04, G3
Leaf basis: image_preimage_def + coord_ops + numeric_evaluation
Taking t=0 gives (1+0, 2+0)=(1,2)=P, and t=1 gives (1+2, 2+4)=(3,6)=Q. So L does pass through the two given points.
S14 · implies
Before
P,Q,R\in L,\ S\notin L
After
P,Q,R\ \text{collinear};\ S\notin L(P,Q)Conditions: none
Rule: collinear_def · Uses: S09, S12, S13
Leaf basis: collinear_def + and_intro_elim
L contains P, Q and R, so by the definition the three points are collinear; S is not on L. Both conclusions hold in this model.
Error controls and negative controls
Appearing to lie on a line is not the same as lying on it
In a sketch S=(2,5) is close to L: at x=2 the point of L is (2,4), only 1 away. The MKO warns against taking a drawing's appearance as a definition. Here membership is decided entirely by the model's equations: the two coordinate equations of S have no common solution, so S∉L however close it looks.
One coordinate is not enough
The first coordinate equation of S, 2=1+2t, has the solution t=1/2, and on its own suggests S is on the line. But t=1/2 in the second coordinate gives 2+4·(1/2)=4≠5. Incidence needs one t that satisfies both coordinate equations (S06 and S12).
Why P≠Q is checked first
If P=Q then Q−P=(0,0) and {P+t·(0,0) ∣ t∈ℝ} is the single point P, not a line. The definition in G2 gives a line only for two distinct points, so S02 must establish P≠Q before S04 can treat L as a line.
- S looks very close to the line in a sketch, so S is on L →
reject — Incidence is decided by equations; the coordinate equations of S demand t=1/2 and t=3/4, with no common solution. - From 2=1+2t we get t=1/2, so S∈L →
reject — The second coordinate fails: 2+4·(1/2)=4≠5. - Any two points P, Q (including P=Q) determine a line {P+t(Q−P)} →
reject — When P=Q this set is the single point P. - In every geometric model, collinearity of three points can be decided by these coordinate equations →
reject — These equations belong to the coordinate model of ℝ²; the MKO says the properties of lines depend on the chosen model, and in another model (for example spherical geometry) lines are defined differently.