Sentence examples for 1 iff from inspiring English sources

p(x) is regular mod x n + 1 iff GCD n, k) = 1.

Therefore, ((mathcal {M},omega ^ models P_{j}(B geq 1) iff ((mathcal {M},omega ^ models P_{i}(psi bowtie alpha ).

Hence, for every ω∈Ω, ((mathcal {M},omega models P_{j}(B geq 1) iff ((mathcal {M},omega models P_{i}(psi bowtie alpha ).

If t ≠ 1, we have r ≤ t (then is called proper and we have t = 1 iff is an ordinary projective plane).

The actor type is captured by a dummy variable θ(a i )={0 1}, which equals 1 iff a i is a governmental actor.

Felicity condition: ∀w"∈ DOXx,w,t [[p]]c,w",S',g,t ⊈ Ct where [[¬p]]c,w,g = 1 iff ∀w'"∈ Ct: [¬p(w'") = 1] & t < t@ ➲ D w) is totally excluded from Ct; hence, the felicity condition is satisfied.

In this domain a DM with CR-preferences weakly prefers LEFT to RIGHT in row r∈{1,…, 2 t+1} iff (1 −σ)[ e+(r−t−1) s] +σ(e+g) ≥e.

The Bounded Hamming Graph, denoted H G Stringss, τ ), is a subgraph of the Hamming Graph where edge (s1, s2) exists iff d ˜ (s 1, s 2 ) ≤ τ.

Then (rho(A)<1) iff (lim _{nlongrightarrowinfty}A^{n}=0).

With divisible goods, a nonrandom quantity q trades with probability 1 and, again, money trades with probability τ where τ<1 iff buyers have sufficient bargaining power.

Q k = -1 iff the belt region contains an orthogonal break path that is not k-barrier covered.