// #5:  N2<=X<=N1  |  N1<=N2 | collapse path (learn that N1=X=N2)
			// #6:  N2<=X<=N1  |  N1<N2  | contradiction
			// #7:  N2<X<N1    |  N1<=N2 | contradiction in the path
			// #8:  N2<X<N1    |  N1<N2  | contradiction

			if strict {
				// Cases #6 and #8: contradiction
				return false
			}

			// We're in case #5 or #7. Try to collapse path, and that will

// efficient.
type factsTable struct {
	// unsat is true if facts contains a contradiction.
	//
	// Note that the factsTable logic is incomplete, so if unsat
	// is false, the assertions in factsTable could be satisfiable
	// *or* unsatisfiable.
	unsat      bool // true if facts contains a contradiction
	unsatDepth int  // number of unsat checkpoints

	facts map[pair]relation // current known set of relation

arrive at the idea of doing those exact calculations. Nowhere is it mentioned
that this test extends the 2-by-1 guess into a 3-by-2 guess. The proof of the
Good Guess Guarantee is only for the 2-by-1 guess and argues by contradiction,
making it difficult to understand how modifications like adding another digit
or adjusting the quotient range affects the overall bound.

All that said, Knuth remains the canonical reference. It is dense but packed

		// This tests for i <= cap, which we can only prove
		// using the derived relation between len and cap.
		// This depends on finding the contradiction, since we
		// don't query this condition directly.
		useSlice(b[:i]) // ERROR "Proved IsSliceInBounds$"
	}
}

func f18(b []int, x int, y uint) {
	_ = b[x]
	_ = b[y]

    if (!isa<func::FuncOp, IfOp, WhileOp>(contraction->getParentOp())) {
      return rewriter.notifyMatchFailure(
          contraction,
          "fused operation must be nested inside a function, If or While");
    }

    // If the contraction is used in multiple places, fusing it will only create
    // more contraction nodes, which is slower.
    if (!contraction.getResult().hasOneUse())

        const int y_col =
            !matmul_op.getAdjY() ? shape_y.back() : *(shape_y.rbegin() + 1);

        // Checks that matrix multiply can perform a valid contraction.
        if (x_col != y_row) {
          result_shape.clear();
          return false;
        }

        result_shape.push_back(x_row);
        result_shape.push_back(y_col);
        return true;

	NewGCM(nonceSize, tagSize int) (cipher.AEAD, error)
}

var _ extraModes = (*aesCipher)(nil)

func NewAESCipher(key []byte) (cipher.Block, error) {
	c := &aesCipher{key: bytes.Clone(key)}
	// Note: 0 is success, contradicting the usual BoringCrypto convention.
	if C._goboringcrypto_AES_set_decrypt_key((*C.uint8_t)(unsafe.Pointer(&c.key[0])), C.uint(8*len(c.key)), &c.dec) != 0 ||

	// We start with the largest factor f = tab[len(tab)-1]
	// that evenly divides q. It does so at most once because
	// otherwise f·f would also divide q. That can't be true
	// because f·f is the next higher table entry, contradicting
	// how f was chosen in the first place.
	// The same reasoning applies to the subsequent factors.
	var p5 uint
	for i := len(tab) - 1; i >= 0; i-- {
		if t, r = t.div(r, q, tab[i]); len(r) == 0 {

- [`KT-65255`](https://youtrack.jetbrains.com/issue/KT-65255) K2 / KJS: "IllegalArgumentException: Candidate is not successful, but system has no contradiction"
- [`KT-65195`](https://youtrack.jetbrains.com/issue/KT-65195) K2: Unexpected exception when executing dynamic array element inc/dec

	//
	// since all the yi for i > 1 are 0 by choice of k: If any of them
	// were > 0, then yh >= b^2 and thus y >= b^2. Then k' = k*2 would
	// be a larger valid threshold contradicting the assumption about k.
	//
	if k < n || m != n {
		tp := getNat(3 * k)
		t := *tp

		// add x0*y1*b
		x0 := x0.norm()
		y1 := y[k:]       // y1 is normalized because y is