Main derivation
S01 · definition
Before
H:\mathbb{R}^2\to\mathbb{R}^2After
H\ \text{isometry}\iff\forall P,Q\in\mathbb{R}^2:\ d(H(P),H(Q))=d(P,Q)Conditions: none
Rule: isometry_def · Uses: G1, isometry_def
Leaf basis: isometry_def + def_unfold
Fix the property to be proved: an isometry is a whole-plane property holding for all pairs of points. To prove a map is an isometry one must argue for an arbitrary pair; to refute it, one counterexample pair is enough. Distance is the Euclidean distance of G1.
S02 · definition
Before
P=(x_1,y_1),\ Q=(x_2,y_2)\ \text{arbitrary}After
\Delta x=x_2-x_1,\ \Delta y=y_2-y_1,\ d(P,Q)=\sqrt{\Delta x^2+\Delta y^2}Conditions: none
Rule: dist_formula · Uses: G1, dist_formula
Leaf basis: def_intro + dist_formula
Take two arbitrary points P and Q, with no condition on them, and name the coordinate differences Δx and Δy. By the distance formula d(P,Q)=√(Δx²+Δy²). Every later argument for a map is carried out for this arbitrary pair, which is what allows generalization at the end.
S03 · equivalent
Before
d(T(P),T(Q))
After
\sqrt{((x_2+t_x)-(x_1+t_x))^2+((y_2+t_y)-(y_1+t_y))^2}=\sqrt{\Delta x^2+\Delta y^2}=d(P,Q)Conditions: none
Rule: dist_formula · Uses: S02, G2
Leaf basis: dist_formula + add_assoc + add_comm + add_inverse + add_zero
Translation: the coordinates of T(Q)−T(P) are (x2+tx)−(x1+tx)=Δx and (y2+ty)−(y1+ty)=Δy (associativity, commutativity and inverses of addition cancel tx and ty; N1 checks this). The radicands are identical, so d(T(P),T(Q))=d(P,Q).
S04 · equivalent
Before
R_{90}(Q)-R_{90}(P)After
(-y_2-(-y_1),\ x_2-x_1)=(-\Delta y,\ \Delta x)
Conditions: none
Rule: distributive · Uses: S02, G2
Leaf basis: distributive + neg_mul + eq_replace
R90(P)=(−y1, x1) and R90(Q)=(−y2, x2). The first coordinate is −y2−(−y1)=−(y2−y1)=−Δy (distributing the sign; N2 checks this), and the second is x2−x1=Δx. The direction is counterclockwise: (1,0) goes to (0,1).
S05 · equivalent
Before
d(R_{90}(P),R_{90}(Q))After
\sqrt{(-\Delta y)^2+\Delta x^2}=\sqrt{\Delta x^2+\Delta y^2}=d(P,Q)Conditions: none
Rule: L_signed_perm · Uses: S04
Leaf basis: neg_mul + add_comm + eq_congruence
The two coordinates of R90(Q)−R90(P)=(−Δy, Δx) are Δy and Δx with a sign and swapped order, so by Lemma L_signed_perm, d(R90(P),R90(Q))=d(P,Q). The lemma is proved in S05.1.
S06 · implies
Before
d(T(P),T(Q))=d(P,Q),\ d(R_{90}(P),R_{90}(Q))=d(P,Q)\ \text{for arbitrary}\ P,QAfter
T\ \text{and}\ R_{90}\ \text{are isometries}Conditions: none
Rule: forall_intro · Uses: S02, S03, S05, S01
Leaf basis: forall_intro + isometry_def
The equalities of S03 and S05 were proved for the arbitrary P and Q of S02, with no condition on them; by generalization they hold for all pairs. By the definition in S01, the translation T and R90 are isometries.
S07 · implies
Before
R_{180},\ R_{270},\ \text{reflections in }x\text{-axis},\ y\text{-axis},\ y=x,\ y=-xAfter
(-\Delta x,-\Delta y),\ (\Delta y,-\Delta x),\ (\Delta x,-\Delta y),\ (-\Delta x,\Delta y),\ (\Delta y,\Delta x),\ (-\Delta y,-\Delta x):\ \text{all isometries}Conditions: none
Rule: L_signed_perm · Uses: S02, G2, S05, S01
Leaf basis: L_signed_perm + forall_intro + distributive + neg_mul
Write H(Q)−H(P) for each: R180 gives (−Δx,−Δy); R270(x,y)=(y,−x) gives (Δy,−Δx); the x-axis reflection gives (Δx,−Δy); the y-axis reflection (−Δx,Δy); the y=x reflection (Δy,Δx); the y=−x reflection M gives (−Δy,−Δx). Each is a signed Δx and Δy, possibly swapped, so by Lemma L_signed_perm and generalization (as in S06) all six maps are isometries; R0 is the identity and is clearly an isometry.
S08 · implies
Before
R_{180}(x,y)=(-x,-y)After
R_{180}\ \text{differs from each of the four axis reflections}Conditions: none
Rule: function_def · Uses: G2, function_def
Leaf basis: function_def + numeric_evaluation + two_nonzero + zero_product
A reflection fixes every point of its axis: the x-axis reflection fixes (1,0), the y-axis reflection (0,1), the y=x reflection (1,1), and the y=−x reflection M fixes (1,−1) (M(1,−1)=(1,−1)). The half-turn has (−x,−y)=(x,y) exactly when 2x=0 and 2y=0, so it fixes only the origin; it sends those four points to (−1,0), (0,−1), (−1,−1), (−1,1), disagreeing with each reflection at that reflection's point. By the definition of equality of functions, R180 is none of these reflections; it is the point inversion, that is the half-turn.
S09 · equivalent
Before
H_c(Q)-H_c(P)
After
(c\Delta x,\ c\Delta y),\quad (c\Delta x)^2+(c\Delta y)^2=c^2(\Delta x^2+\Delta y^2)
Conditions: none
Rule: distributive · Uses: S02, G2
Leaf basis: distributive + power_laws + mul_comm
Hc(Q)−Hc(P)=(cx2−cx1, cy2−cy1)=(cΔx, cΔy) (distributivity). The sum of squares is c²Δx²+c²Δy²=c²(Δx²+Δy²) ((ab)²=a²b² and distributivity; N4 checks this).
S10 · equivalent
Before
d(H_c(P),H_c(Q))=\sqrt{c^2(\Delta x^2+\Delta y^2)}After
|c|\sqrt{\Delta x^2+\Delta y^2}=|c|\,d(P,Q)Conditions: S=\Delta x^2+\Delta y^2\ge 0
Rule: L_sqrt_scale · Uses: S09, S02
Leaf basis: sqrt_def + abs_rules + power_laws + order_basics + square_nonneg
Let S=Δx²+Δy²≥0 (a sum of squares is nonnegative). By Lemma L_sqrt_scale, √(c²S)=|c|√S, so d(Hc(P),Hc(Q))=|c|d(P,Q). The factor is |c|, not c: a distance is never negative. The lemma is proved in S10.1 to S10.3.
S11 · implies
Before
c=-1:\ d(H_{-1}(P),H_{-1}(Q))=|-1|\,d(P,Q)After
H_{-1}=R_{180}\ \text{is an isometry}Conditions: none
Rule: forall_intro · Uses: S10, G2
Leaf basis: abs_rules + forall_intro + numeric_evaluation
|−1|=1, so distance is unchanged for arbitrary P and Q and H(−1) is an isometry; indeed H(−1)(x,y)=(−x,−y) is exactly the half-turn R180, consistent with S07. This shows that a negative coefficient need not destroy isometry.
S12 · implies
Before
c=-2,\ P=(3,1),\ Q=(-1,4)
After
d(P,Q)=5,\ d(H_{-2}(P),H_{-2}(Q))=10=|-2|\cdot 5Conditions: none
Rule: numeric_evaluation · Uses: S10, G3
Leaf basis: numeric_evaluation + dist_formula
Numerical check: Q−P=(−4,3), d(P,Q)=√(16+9)=5. H(−2)(P)=(−6,−2) and H(−2)(Q)=(2,−8), with difference (8,−6) and distance √(64+36)=10=2·5. The scale is |−2|=2, as S10 says; it is not −2, and the distance is not −10. This is one example; the general statement comes from S10.
S13 · implies
Before
c=2:\ H_2(0,0)=(0,0),\ d(H_2(P),H_2(Q))=10\neq 5=d(P,Q)
After
H_2\ \text{is not an isometry}Conditions: none
Rule: quantifier_negation · Uses: S12, S01
Leaf basis: quantifier_negation + isometry_def + numeric_evaluation
|2|=|−2|, so the same pair P, Q has image distance 10≠5 under H2 too. Isometry requires every pair, and one counterexample refutes it: ¬∀ is ∃¬. Although H2 fixes the origin (a fixed point), that cannot establish whole-plane isometry, which is exactly the warning of MKO assumption a3.
S14 · implies
Before
c=0:\ H_0(x,y)=(0,0)
After
d(H_0(P),H_0(Q))=0\neq 5:\ \text{degenerate constant map, not an isometry}Conditions: none
Rule: quantifier_negation · Uses: S10, G3, S12
Leaf basis: quantifier_negation + numeric_evaluation
With c=0 every point goes to the origin; the map is constant. For the distinct P and Q, d(P,Q)=5>0 but the image distance is 0. The factor |0|=0 is not positive, so the map is neither an isometry nor a positive-scale similarity in the MKO's sense; it is degenerate.
S15 · case_split
Before
d(H_c(P),H_c(Q))=|c|\,d(P,Q)\ \forall P,Q
After
c=0:\ \text{degenerate};\ |c|=1:\ \text{isometry};\ \text{other }c\neq 0:\ \text{similarity of scale }|c|,\ \text{not an isometry}Conditions: none
Rule: case_analysis · Uses: S10, S11, S13, S14, S06, S07
Leaf basis: case_analysis + abs_rules + quantifier_negation + forall_intro
Three cases cover every real c. c=0: S14, a degenerate constant map. |c|=1 (c=1 the identity, c=−1 the half-turn): S10 gives unchanged distances, an isometry. Any other nonzero c: S10 multiplies every distance by the positive |c|, so the map is a similarity of scale |c|; and as |c|≠1, the pair P=(0,0), Q=(1,0) has d(P,Q)=1 but image distance |c|≠1, so it is not an isometry. This agrees with the MKO's classification.