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View the oriented square lattice as $(x,n)$ with $x+n$ even, and arrows $(x,n)\to (x\pm 1,n+1)$. Each graph vertex has two incoming bonds. Start with independent oriented bond percolation of parameter $p$, and declare a site open when at least one of its incoming bonds is open. Its probability of being open isIncoming bond sets for different heads are disjoint, so these site indicators are independent: they are precisely an oriented site percolation configuration with parameter $r$.

Every graph vertex after the first on an open oriented bond path is open as a site, because its incoming path bond is open. The root's two incoming bonds belong to earlier time layers and are independent of all bonds used by a forward path from the root. Thus the event that the root site is open, of probability $r$, is independent of bond survival from that root. On their intersection the entire infinite bond path is also an infinite open site path. Hence, writing ${\theta }_b,{\theta }_s$ for the two survival probabilities,This is the incoming-edge coupling of site and bond percolation. If site survival is defined conditional on an open root, the extra factor is omitted; the critical probability is the same. If the model is drawn only in a time half-plane, one can independently sample the root's incoming bonds as auxiliary variables.

For every $p>p_c(\mathrm{bond})$, bond survival is positive by monotonicity and the definition of the critical probability. Therefore $p_c(\mathrm{site})\le 1-(1-p{)}^2$. Let $p$ decrease to the bond critical probability and use continuity of this polynomial, obtainingIf the bond critical probability equals one, the inequality is trivial; no assertion about survival exactly at criticality is used.