When I execute f[g[a, b]] /. f[g[l__]] :> f[l] I get f[a, b] as expected.
When I execute f[a + b] /. f[Plus[l__]] :> f[l] I get f[a + b]. (I expect f[a, b] again)
Why? (Wolfram Mathematica 12.2.0)
$\begingroup$
$\endgroup$
1
1 Answer
$\begingroup$
$\endgroup$
2
This happens because Plus[l__] evaluates to l__ before the pattern is even replaced. Try instead:
f[a + b] /. f[HoldPattern@Plus[l__]] :> f[l]
answered Mar 8, 2021 at 13:02
-
1$\begingroup$ you can also do
f[a + b] /. f[l_Plus]] :> f[l]$\endgroup$ Commented Mar 8, 2021 at 13:16 -
2$\begingroup$ That still evaluates to
f[a + b], right? $\endgroup$ Commented Mar 8, 2021 at 13:20
f[Plus[a, b]] /. f[g_[l1__]] :> f[l1]gives:f[a, b]$\endgroup$