From 9c1bed84b1fb1a6a3012f23dccf867491cbe960b Mon Sep 17 00:00:00 2001 From: Hykilpikonna Date: Mon, 8 Nov 2021 21:11:14 -0500 Subject: [PATCH] [F] A4 P2.3.a Change wording --- assignments/a4/a4.tex | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/assignments/a4/a4.tex b/assignments/a4/a4.tex index caacd6c..ad706cc 100644 --- a/assignments/a4/a4.tex +++ b/assignments/a4/a4.tex @@ -105,7 +105,7 @@ Prove that each loop invariant holds. \begin{proof} : \\ Variables: In this proof, $N$ is the abbreviation for the list \texttt{nums\_so\_far}. \\ \\ -Assumption 1: The loop invariant 1 is true for the last iteration. \\ +Assumption 1: The loop invariant 1 is true for the previous iteration. \\ That is $\forall k_2 \in N, gcd(k_2, 2) = 1 \land gcd(k_2, 3) = 1$ \\ \\ Assumption 2: The statement proven in Part 1.1: \\ @@ -115,7 +115,7 @@ We need to prove: $\forall k \in N \cup \{ N[-2] + 6 \}, gcd(k, 2) = 1 \land gcd Which is equivalent to: $\forall k \in N, gcd(k, 2) = 1 \land gcd(k, 3) = 1$ and \\ $gcd(N[-2] + 6, 2) = 1 \land gcd(N[-2] + 6, 3) = 1$ \\ \\ -Since the first part is the same as the last iteration, it is true. \\ +Since the first part is the same as the previous iteration, it is true. \\ What we need to prove becomes: $gcd(N[-2] + 6, 2) = 1 \land gcd(N[-2] + 6, 3) = 1$ \\ \\