By Fubini the plane integral of g(x)h(y) over any rectangle factors as the product of the two coordinate integrals. Additivity of the plane integral over the tiling — valid since distinct tiles meet only in their measure-zero boundaries — writes \int\!\!\int_R g\,h as the sum over tiles of \int\!\!\int_{T_i} g\,h. Each summand is a product with a vanishing factor, so it is 0; hence \bigl(\int_R g\bigr)\bigl(\int_R h\bigr) = 0, and a product of reals (or complex numbers) vanishes only if a factor does.

Take g(x) = h(x) = e^{2\pi i x}. The one-dimensional integral \int_a^b e^{2\pi i x}\,dx = (e^{2\pi i b} - e^{2\pi i a})/(2\pi i) vanishes if and only if e^{2\pi i(b-a)} = 1, i.e. b - a \in \mathbb{Z}. So a tile's coordinate integral vanishes exactly when that side is an integer; the hypothesis gives the per-tile dichotomy, the engine Lemma 7.2 transports it to R, and the same criterion reads off an integer side. The complex exponential is what makes the criterion an exact "integer side" statement, with no reflected solutions.

Use the real integrand \sin(2\pi(x - R_x))\sin(2\pi(y - R_y)), where (R_x, R_y) is the corner of R; shifting the integrand by the corner replaces Wagon's "place R in standard position". The factor integral is \int_a^b \sin(2\pi(x-s))\,dx = (\cos 2\pi(a-s) - \cos 2\pi(b-s))/(2\pi). It vanishes whenever b - a \in \mathbb{Z} (used for the tiles, where the shift is irrelevant); and at the corner, where the lower limit equals the shift, it reduces to (1 - \cos 2\pi(b-s))/(2\pi), which vanishes if and only if b - s \in \mathbb{Z}. The engine Lemma 7.2 then forces an integer side of R.

Colour the plane in a \tfrac12 \times \tfrac12 checkerboard with a corner at R's corner; "equal black and white" over a region means \int\!\!\int (-1)^{\lfloor 2x\rfloor}(-1)^{\lfloor 2y\rfloor} = 0. The one-dimensional factor (-1)^{\lfloor 2(x-s)\rfloor} is the \pm 1 square wave of period 1; its integral over any integer-length interval is 0, so every tile with an integer side is balanced. By the engine Lemma 7.2 so is R. But over [s, b] the square wave integrates to the triangle wave \min(r, 1-r), with r the fractional part of b - s, which is nonzero unless b - s \in \mathbb{Z}; hence R has an integer side.

Floor and ceiling agree at the integers and differ by one everywhere else, so c(x) is 2\lfloor x - a\rfloor in the first case and 2\lfloor x - a\rfloor + 1 in the second.

That number is by definition the fg-area of the pair of grid coordinate functions, so this is fg-area additivity Lemma 7.33, read back in the integers. It is what Wagon gets from the auxiliary tiling, here without having to check that the translated tiles tile the translated rectangle.

Base the half-unit grid at the lower-left corner of R. The endpoints of a side of integer length n are n apart, so the translation moves them equally and the increment of the grid coordinate across that side is 2n: every tile covers an even number of squares, and hence Lemma 7.7 so does R. The corner of R has grid coordinate 0, so that count is the product of the grid coordinates of the far edges of R, and one of the two factors is even — which says Lemma 7.6 that the corresponding side of R has integer length. (Wagon phrases the last step as a contradiction: a translated rectangle with no integer side has both sides equal to half an odd integer, hence an odd number of squares.)

Let \varepsilon_x(u) be 0 when u - R_x \in \mathbb{Z} and 1 otherwise, and similarly define \varepsilon_y; perturb the coordinates by \varphi_t(u) = u + t\varepsilon_x(u) and \psi_t(v) = v + t\varepsilon_y(v). The perturbed area of a rectangle S is then the value at t of the polynomial P_S = (\Delta\varepsilon_x X + w)(\Delta\varepsilon_y X + h), with w, h the sides of S and \Delta\varepsilon the indicator increments across them; its quadratic coefficient is the fg-area of the indicator pair. If a tile has an integer side, the two endpoints of that side have the same indicator, the corresponding factor is constant, and the quadratic coefficient vanishes: the tile's perturbed area is linear or constant in t, as in Wagon's text. Additivity of the fg-area Lemma 7.33 equates \sum_i P_{T_i} with P_R at every real t — Wagon needs t small so that the moved segments still bound a tiling, the algebraic identity does not — hence \sum_i P_{T_i} = P_R as polynomials. If neither side of R is an integer, each lower endpoint has indicator 0 and each upper endpoint 1, so the quadratic coefficient of P_R is 1; comparing X^2-coefficients yields 0 = 1, a contradiction. (A polynomial-free variant of the same computation: each tile's term is affine in t, so the second finite difference F(2) - 2F(1) + F(0) of the identity vanishes tile by tile, while it equals 2 on (w + t)(h + t).)

Disjointness: a point common to two half-open cells moves down-and-left — to the midpoint between the higher of the two lower-left corners and the point itself — into the interiors of both tiles, contradicting interior-disjointness. Covering: a point of the ambient half-open cell is approached from below-left; every such nudge stays in R, hence in some tile, and since there are finitely many tiles one tile contains nudges arbitrarily close in. That tile is closed, so it contains the point, and the nudges witness the two strict inequalities of its half-open cell.

A lattice point (a/p, b/p) lies in the half-open cell iff \lfloor p x_0\rfloor < a \le \lfloor p x_1\rfloor and likewise for b, which gives the product formula; additivity is then exactly the exact-partition property Lemma 7.10 of the half-open cells, read through this membership description.

A tile side of integer length n contributes the floor-difference factor \lfloor p x_0 + pn\rfloor - \lfloor p x_0\rfloor = pn, so every tile's lattice count is divisible by p, and by additivity Lemma 7.11 so is the count of R, a product of two floor differences. Primality forces p to divide one factor, say \lfloor p x_1\rfloor - \lfloor p x_0\rfloor = pm; then p(x_1 - x_0) - pm is a difference of two floor remainders, each in [0,1), so |width - m| < 1/p.

Suppose neither side of R is an integer. Each of the width and the height then keeps some fixed positive distance from every integer — at least the smaller of its fractional part and one minus it. Choosing a prime p larger than the reciprocals of both distances (there are infinitely many primes), no side of R can be within 1/p of an integer, contradicting the claim Lemma 7.12 for this p.

Fix a point (u, v) and take for f and g the indicator functions of \{u\} and \{v\}. The fg-area of a rectangle is then (\mathbb{1}[x_1 = u] - \mathbb{1}[x_0 = u])(\mathbb{1}[y_1 = v] - \mathbb{1}[y_0 = v]), the number of corners at (u, v) with the left and bottom edges counted negatively. Being an fg-area it is additive over the tiling Lemma 7.33, and modulo 2 subtraction and addition agree, so each signed count may be replaced by the corner count.

Count the incidences between the points of Z and the tiles having them as a corner. Tile by tile the total is even by hypothesis. Counting the same incidences point by point instead, and replacing each point's count by the corresponding count for R Lemma 7.14, leaves the parity unchanged, so the number of corners of R in Z is even too.

Take for Z the connected component of the lower-left corner of R, that is, the vertices a walk starting there reaches. An edge of \Gamma has both of its ends in Z or neither, and the sides that a tile with an integer side contributes to \Gamma pair up its four corners; so every tile has an even number of corners in Z, and by the double count Lemma 7.15 so has R. The lower-left corner of R is one of them, hence not the only one.

An edge of \Gamma is a side of a tile, parallel to an axis and of integer length, so it leaves the pair of fractional parts of a point unchanged, and hence so does a walk. The walk to a second corner of R Lemma 7.16 therefore preserves the fractional part of the abscissa or of the ordinate — of both, if it ends at the opposite corner — which says that the width or the height of R is an integer.

The number of corners a rectangle has on the grid is the number of its vertical edges over X times the number of its horizontal edges over Y, so by hypothesis every tile has an even number of them, and the double count Lemma 7.15 applies.

Take for X and Y the points of the lattices through the left and bottom edges of R that lie inside R. A tile of integer width has its two vertical edges an integer apart, so both or neither lie on the lattice, and likewise in the other direction; the hypothesis of the count Lemma 7.18 therefore holds, and R has evenly many corners on the grid. Its lower-left corner is one of them. If neither side of R were an integer, its right edge would miss X and its top edge would miss Y, leaving that corner as the only one — an odd number.

Discard the degenerate tiles and cut each remaining tile into unit pieces along a side of integer length, of which it has at least one, and which is at least one unit long since the tile is nondegenerate. The half-open cells of the pieces of a tile partition its own, so the cells of all the pieces still partition those of R.

Induct on the number of unit levels the strip has risen. Nothing crosses the top edge of the H-tile inside the strip, since the only tile below that edge there is the H-tile itself. If nothing crosses the height reached so far and no H-tile starts there, then the tile above each point of that height starts exactly there, hence is a V-tile and is one unit tall; it therefore spans the whole level, nothing crosses the next height, and the strip is still inside R. Each level raises the strip by a unit, so it cannot rise forever.

Walk down from the given H-tile: while the column below the current H-tile is blocked, step onto the H-tile blocking it, which lies strictly lower, so the walk stops at an H-tile whose column reaches the bottom of R. From there build the staircase upwards Lemma 7.21 column by column, each step raising the top edge to a strictly higher one of the finitely many heights carrying a horizontal edge of the tiling. A tile meeting a column lies inside it, so it meets no other column, and the strip has a single abscissa along it; and consecutive columns overlap horizontally, since the H-tile carrying the upper column crosses the lower one. A tile to the left of the staircase at one height and to its right at another would have to fit in the gap between two consecutive columns, and there is no gap.

Each tile is cut at a single abscissa: its right edge if it lies left of the strip, its left edge less one if it lies right of the strip, and the left edge of the strip if it lies inside it — and since the staircase is a wall Lemma 7.22 the same alternative holds at all of that tile's heights. A point of the shrunken rectangle left of the strip comes from the point itself and one to its right from its translate one unit rightwards, so the half-open cells of the pieces partition those of the shrunken rectangle.

Normalize the tiling Lemma 7.20 and induct on the number of its H-tiles. With no H-tile every tile has integer height, hence so does R. Otherwise grow the staircase from an H-tile Lemma 7.22. If R is narrower than two units then the H-tile leaves no room beside it and R is exactly one unit wide. Otherwise cut the staircase out Lemma 7.23: heights never change, so V-tiles stay V-tiles, and an H-tile is carried over whole or swallowed by the strip, so the new tiling is normalized and has fewer H-tiles — the one the staircase was grown from is gone. Its rectangle is one unit narrower and just as tall, so an integer side of it is an integer side of R.

Approach the point of the link at that height from below left; the tile whose half-open cell contains it has the line on or to the right of its own right edge. It cannot reach past the line, for a tile crossing the line would block a height interior to the link, and it cannot reach past either end of the link, because whatever blocks that end — a tile crossing the line, or a horizontal edge arriving at the line from both sides — would share a point with it.

The tiles on either side of the link cut it into consecutive pieces Lemma 7.25, so the strip of width w that the tiles on the left sweep out is exactly the strip that the tiles on the right vacate; no other tile can reach into it, and the half-open cells of the new family therefore still partition those of R.

Suppose not. Then from any H-tile one can step to an H-tile directly above or below across an H-link, so the H-tiles reach every height of R; and from any V-tile one can step to a V-tile across a V-link on either side, so the V-tiles reach from the left edge of R to its right edge. Follow the V-tiles rightwards, carrying the assertion that every H-tile overlapping the current V-tile in height lies to the right of it. It holds at the left edge, where there is no room on the left. It survives a step: an H-tile lying on the left of the V-link just crossed can be walked up and down along H-links to the height of the previous V-tile, staying on the left throughout, since a V-link blocks every H-link it meets and the H-tiles bordering an H-link do not reach past its ends — contradicting the assertion for the previous V-tile. But at the right edge of R the assertion is absurd, since the H-tiles reach that height too.

Induct on the number of tiles. Degenerate tiles have empty half-open cells and may be discarded. If every tile has integer width the fg-area for the fractional part horizontally and the identity vertically vanishes on every tile, hence on R, so R has integer width; likewise if every tile has integer height. Otherwise some link is reducible Lemma 7.27, say a V-link with only H-tiles on its right; take for w the width of the narrowest of them and push Lemma 7.26. Heights never change, so V-tiles stay V-tiles, and widths change by the integer w, so H-tiles stay H-tiles; but the narrowest tile on the right is squeezed to nothing, so the new tiling of R has fewer tiles and the induction hypothesis applies. The three other ways a link can be reducible are this one read in the mirror image of the tiling, in its transpose, and in the mirror image of its transpose.