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Summary

I am struggling with substitution rules.

Example

Here are several cases which are problematic:

Clear[a,α];
{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a + b -> α}

Actual result:

{1 + a + 2 b, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}

Desired result:

{1 + α +  b, -α, 2 α, α^2}

Question

Currently, the rule is permuted for every case, e.g.

2 a + 2 b -> 2α

Can the alpha substitution rule be generalized That, is there a single rule to handle all cases?

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    $\begingroup$ You mean: "One to rule them all"? ;) $\endgroup$ – Henrik Schumacher Nov 4 '18 at 18:38
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    $\begingroup$ Use PolynomialReduce to obtain algebraic "substitutions". In[208]:= PolynomialReduce[{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2}, a + b - alpha, {a, b}][[All, 2]] Out[208]= {1 + alpha + b, -alpha, 2 alpha, alpha^2} $\endgroup$ – Daniel Lichtblau Nov 4 '18 at 19:14
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    $\begingroup$ Generally, to apply a relationship broadly, write the corresponding rule such that the LHS of the rule is as simple as possible, e.g., solution posted by @HenrikSchumacher. Since rules are applied to the structure of the internal (FullForm) representation, this will result in the highest number of matches with the LHS. $\endgroup$ – Bob Hanlon Nov 4 '18 at 19:38
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{a + 2 b + 1, -a - b, 2 a + 2 b, a^2 + 2 a b + b^2} /. {a -> α - b} // Simplify

{1 + b + α, -α, 2 α, α^2}

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Also

Simplify[{a + 2 b + 1, -a - b, 2 a + 2 b,  a^2 + 2 a b + b^2}, {a + b == α}]

{1 + b + α, -α, 2 α, α^2}

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