diff --git a/Electronics/Digital_Circuits/Half_adder_and_full_adder.md b/Electronics/Digital_Circuits/Half_adder_and_full_adder.md
index d7cdb29..fa13907 100644
--- a/Electronics/Digital_Circuits/Half_adder_and_full_adder.md
+++ b/Electronics/Digital_Circuits/Half_adder_and_full_adder.md
@@ -87,9 +87,11 @@ And the carry-out bit replicates the truth conditions of [AND](/Hardware/Logic_G
It is therefore possible to implement a half-adder with just these two logic gates:
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The digital circuit above has the same inputs and outputs as the half adder diagram above.
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## Full adder
@@ -126,4 +128,4 @@ The sume of HA1 ($0 + 0$) is passed in to the B input on HA2 and the $1$ is pass
At this point we have completed the addition and have successfully added the three bits: $1$, $0$, and $0$ to get $1$ as ths sum. However we also have to account for the fact that the addition may result in its own carry-out bit. What if the inputs were $1, 1, 0$ for example?
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