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In Boolean algebra, the product $yz$ is the consensus of $xy$ and $\bar{x}z$. If $yz=1$, either $x=1$ and $xy=1$, or $x=0$ and $\bar{x}z=1$, so adding the consensus does not change the Boolean function. It can change a circuit's transient behavior: while $x$ switches, the two original product terms may briefly both be zero because of unequal propagation delays, but $yz$ remains one. Keeping the logically redundant consensus product can therefore remove a static-one logic hazard.

The internal Boolean functions are $x=\bar{a}\bar{c}$, $y=bc$, $z=\bar{c}d$, $w=y+z$ and $q=x+w$. Put the falling edge of $c$ at $t=0$. Since $a=1$ throughout, $x=0$ throughout. The remaining transitions, including every stated propagation delay, areThe inverter delays the rise of $\bar{c}$ by ten nanoseconds, and the $z$ AND gate adds another ten. Consequently $y$ and $z$ are simultaneously zero for ten nanoseconds. This gap propagates through the two OR gates. The output has a static-one logic hazard: a low glitch on $30\le t<40$ ns, despite its equal initial and final logical values.

A logic hazard in which an output intended to remain one briefly falls to zero. In a sum of products, cover every adjacent pair of one-cells in a Karnaugh map with a common product implicant. That term omits the changing input and remains one throughout the transition, holding the OR output high regardless of the other path delays. This argument assumes one input changes at a time and the circuit starts settled.

For a concrete circuit, let $x=\bar{a}\bar{c}$, $y=bc$, $z=\bar{c}d$, $w=y+z$ and $q=x+w$. Give every gate, including each inverter, propagation delay 10 ns. Starting from $a=b=c=d=1$, lower $c$ at time zero. Signal $y$ falls at 10 ns, while $z$ rises only at 20 ns. Their gap passes through the two OR gates, making $q$ fall during 30–40 ns even though its steady-state value remains one.

Adding the two consensus theorem products $\bar{a}b$ and $bd$ preserves the Boolean function and holds the output high across every potentially hazardous single-input transition. The resulting five-term expression factors as $q=bc+(\bar{a}+d)(b+\bar{c})$. When a change of $c$ could leave the output steadily one, $b=1$ and $\bar{a}+d=1$ keep the second product steadily one. In the zero-to-zero case with changing $c$, $b=0$ and $\bar{a}+d=0$ keep both output products zero. Changes of $a,b,d$ use only one polarity each and therefore have no opposite-polarity race. This factored implementation uses three OR gates, two AND gates and two NOT gates, with a 40 ns maximum input-to-output path at the stated delays.