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If I run the following code:

Do[ Print["*** ", HoldForm[x] /. x -> a, " ***"]; Print["Replace: ", (HoldForm[x] /. x -> a) /. {n1 -> 1, n2 -> 2, n3 -> 3}]; Print["FullForm: ", FullForm[a]]; , {a, {n1 n2, -n1 n2, n1 n2 n3}} ]

I get

 *** n1 n2 *** Replace: 1 x 2 FullForm: Times[n1,n2] *** -n1 n2 *** Replace: -2 FullForm: Times[-1,n1,n2] *** n1 n2 n3 *** Replace: 2 x 3 FullForm: Times[n1,n2,n3]

I kind of understand what is going on, but I would like the output in the second case to be -1 x 2 and in the third case 1 x 2 x 3. How do I do that?

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    $\begingroup$ Because $1$ is an identity element for multiplication, it gets removed at once in Times[]. (Similarly, $0$ is quickly removed in Plus[].) Perhaps you want something like Inactivate[Times[1, 2, 3], Times]? $\endgroup$ Commented May 14, 2020 at 15:31
  • $\begingroup$ Thank you. Unfortunately, I need to block 1, not Times, and Inactivate[1] is just the same as 1. However, your solution inspired me to find a solution of my own, which I will post now. $\endgroup$ Commented May 14, 2020 at 19:26

1 Answer 1

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It is enough to write the substitution rule as

{n1 -> HoldForm[1], n2 -> 2, n3 -> 3}

If later one wants the expression evaluated, it will be enough to use HoldForm[] twice.

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