Main derivation
S01 · definition
Before
\operatorname{area}([x_0,x_1)\times[y_0,y_1))After
(x_1-x_0)(y_1-y_0)\quad(x_0\le x_1,\ y_0\le y_1)
Conditions: x_0\le x_1; y_0\le y_1
Rule: mko_area_def · Uses: G1
Leaf basis: mko_area_def + def_unfold
State the MKO's restricted model: a half-open rectangle with ordered endpoints has area width times height; a unit cell has area one, and a finite union of distinct unit cells has area equal to their number; the area of the union of two rectangles is the sum of their areas minus the area of their intersection. Area is measured in square units and is not the perimeter.
S02 · equivalent
Before
\operatorname{area}(A)After
(4-0)(3-0)=12
Conditions: 0\le 4; 0\le 3
Rule: mko_area_def · Uses: S01, G2
Leaf basis: mko_area_def + numeric_evaluation
A=[0,4)×[0,3) has ordered endpoints (0≤4, 0≤3), so the formula of S01 applies: width 4−0=4, height 3−0=3, area 4·3=12 square units.
S03 · equivalent
Before
\operatorname{area}(B)After
(6-2)(5-1)=16
Conditions: 2\le 6; 1\le 5
Rule: mko_area_def · Uses: S01, G2
Leaf basis: mko_area_def + numeric_evaluation
B=[2,6)×[1,5): width 6−2=4, height 5−1=4, area 16 square units. The width is the right endpoint minus the left endpoint, not the right endpoint itself.
S04 · equivalent
Before
[0,4)\cap[2,6)
After
[\max(0,2),\ \min(4,6))=[2,4)
Conditions: none
Rule: L_interval_cap · Uses: G2, L_interval_cap
Leaf basis: L_interval_cap + numeric_evaluation
By Lemma L_interval_cap (a₀=0, a₁=4, b₀=2, b₁=6) the intersection of the two x-intervals is [max(0,2), min(4,6))=[2,4). The lemma is proved in S04.1 to S04.3.
S05 · equivalent
Before
[0,3)\cap[1,5)
After
[\max(0,1),\ \min(3,5))=[1,3)
Conditions: none
Rule: L_interval_cap · Uses: G2, L_interval_cap
Leaf basis: L_interval_cap + numeric_evaluation
Apply the same lemma to the y-intervals (a₀=0, a₁=3, b₀=1, b₁=5): the intersection is [1,3).
S06 · equivalent
Before
A\cap B
After
([0,4)\cap[2,6))\times([0,3)\cap[1,5))=[2,4)\times[1,3)
Conditions: none
Rule: cartesian_product_def · Uses: S04, S05, G2
Leaf basis: cartesian_product_def + set_operations_def + and_intro_elim
(x,y)∈A∩B ⇔ (x∈[0,4) and y∈[0,3)) and (x∈[2,6) and y∈[1,5)) ⇔ x∈[0,4)∩[2,6) and y∈[0,3)∩[1,5). So the intersection of two rectangles is again a rectangle whose sides are the intersections of the corresponding intervals; substituting S04 and S05 gives [2,4)×[1,3).
S07 · equivalent
Before
\operatorname{area}(A\cap B)After
\max(0,4-2)\cdot\max(0,3-1)=2\cdot 2=4
Conditions: 2\le 4; 1\le 3
Rule: mko_area_def · Uses: S01, S06
Leaf basis: mko_area_def + max_min_def + numeric_evaluation
By MKO assumption a2 the width of the intersection is max(0, min(upper endpoints)−max(lower endpoints))=max(0,4−2)=2 and the height is max(0,3−1)=2. The outer max(0, ·) makes the width 0 rather than negative when the intersection is empty. Here 2≤4 and 1≤3, so [2,4)×[1,3) is a rectangle with ordered endpoints and area 4.
S08 · theorem_application
Before
\operatorname{area}(A\cup B)After
12+16-4=24
Conditions: none
Rule: mko_area_def · Uses: S02, S03, S07, mko_area_def
Leaf basis: mko_area_def + numeric_evaluation
By the MKO's inclusion and exclusion for two rectangles: area(A∪B)=area(A)+area(B)−area(A∩B)=12+16−4=24. The intersection is counted in both A and B, so it is subtracted once; adding directly to get 28 would count it twice.
S09 · equivalent
Before
A,\ B
After
A=\bigcup_{0\le i\le 3,\ 0\le j\le 2}Q_{i,j},\quad B=\bigcup_{2\le i\le 5,\ 1\le j\le 4}Q_{i,j}Conditions: none
Rule: cell_decomp · Uses: G1, G2, cell_decomp
Leaf basis: cell_decomp + cartesian_product_def + finite_counting_basics
By cell_decomp, [0,4) is the union of [i,i+1) for i=0,1,2,3 and [0,3) is the union of [j,j+1) for j=0,1,2; the Cartesian product of two unions is the union of the products of the pairs, so A is the union of 4·3=12 unit cells Q_{i,j}. Likewise B is the union of the 16 cells with i∈{2,3,4,5} and j∈{1,2,3,4}. The half-open convention makes distinct cells pairwise disjoint: each point lies only in the cell (⌊x⌋,⌊y⌋).
S10 · equivalent
Before
A\cup B
After
\bigcup_{(i,j)\in I_A\cup I_B}Q_{i,j}Conditions: none
Rule: set_operations_def · Uses: S09, set_operations_def
Leaf basis: set_operations_def + set_extensionality
Let I_A and I_B be the index sets of the cells of A and B. A point is in A∪B ⇔ it is in a cell with (i,j)∈I_A or in one with (i,j)∈I_B ⇔ it is in a cell with (i,j)∈I_A∪I_B. Indices are merged as a set, so a cell appearing on both sides (for example Q_{2,1}) counts once (MKO assumption a3: deduplicate by coordinates).
S11 · equivalent
Before
|I_A\cup I_B|
After
4+6+6+4+4=24
Conditions: none
Rule: finite_counting_basics · Uses: S10, S09, finite_counting_basics, mko_area_def
Leaf basis: finite_counting_basics + mko_area_def + numeric_evaluation
Count the distinct i row by row (fixed j): for j=0 only A contributes, i∈{0,1,2,3}, 4 cells; for j=1, A gives {0,1,2,3} and B gives {2,3,4,5}, merging to {0,1,2,3,4,5}, 6 cells; j=2 likewise 6 cells; j=3 only B, {2,3,4,5}, 4 cells; j=4 also 4 cells. Different rows are disjoint, so the counts add (finite_counting_basics) to 24. By the MKO the area of a union of distinct unit cells is their number, so area(A∪B)=24, in agreement with S08.
S12 · implies
Before
A\cap C,\ C=[4,7)\times[0,3)
After
[\max(0,4),\min(4,7))=[4,4)=\varnothing,\ \operatorname{area}(A\cup C)=12+9-0=21Conditions: none
Rule: L_interval_cap · Uses: G3, L_interval_cap, S01, S02
Leaf basis: L_interval_cap + mko_area_def + numeric_evaluation
In the x-direction: [0,4)∩[4,7)=[max(0,4), min(4,7))=[4,4), and no x satisfies 4≤x<4, so it is empty; hence A∩C=∅, width max(0,4−4)=0 and intersection area 0. The area of C is (7−4)(3−0)=9, so area(A∪C)=12+9−0=21. For example (4,1)∈C but 4∉[0,4), so (4,1)∉A: two half-open rectangles that merely touch share no point.
S13 · implies
Before
Z=[3,3)\times[0,5)
After
Z=\varnothing,\ \operatorname{area}(Z)=(3-3)(5-0)=0,\ \operatorname{area}(A\cup Z)=12Conditions: 3\le 3
Rule: mko_area_def · Uses: G1, G3, S01, S02
Leaf basis: mko_area_def + mul_zero + set_operations_def
By the half-open convention, [3,3) needs 3≤x<3, which no x satisfies, so [3,3)=∅ and Z=∅×[0,5)=∅. The formula also gives (3−3)(5−0)=0·5=0 (mul_zero). A∪Z=A∪∅=A, whose area is still 12. A height of 5 does not give the empty rectangle any area (MKO: zero width or height makes the rectangle empty, with area zero).
S14 · implies
Before
\text{tasks of the goal}After
\operatorname{area}(A\cup B)=24,\ \operatorname{area}(A\cup C)=21,\ \operatorname{area}(Z)=0Conditions: none
Rule: and_intro_elim · Uses: S02, S03, S07, S08, S11, S12, S13
Leaf basis: and_intro_elim
Combine: area(A)=12, area(B)=16, A∩B=[2,4)×[1,3) with area 4, area(A∪B)=24 (inclusion and exclusion and the cell count agree), A∩C=∅ and area(A∪C)=21, area(Z)=0. Every area value is in square units.