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    <p>You can follow the following steps to reduce lambda expressions:</p> <ol> <li>Fully parenthesize the expression to avoid mistakes and make it more obvious where function application takes place.</li> <li>Find a function application, i.e. find an occurrence of the pattern <code>(λX. e1) e2</code> where <code>X</code> can be any valid identifier and <code>e1</code> and <code>e2</code> can be any valid expressions.</li> <li>Apply the function by replacing <code>(λx. e1) e2</code> with <code>e1'</code> where <code>e1'</code> is the result of replacing each free occurrence of <code>x</code> in <code>e1</code> with <code>e2</code>.</li> <li>Repeat 2 and 3 until the pattern no longer occurs. Note that this can lead to an infinite loop for non-normalizing expressions, so you should stop after 1000 iterations or so ;-)</li> </ol> <p>So for your example we start with the expression</p> <pre><code>((λm. (λn. (λa. (λb. (m ((n a) b)) b)))) (λf. (λx. x))) (λf. (λx. (f x))) </code></pre> <p>Here the subexpression <code>(λm. (λn. (λa. (λb. (m ((n a) b)) b)))) (λf. (λx. x))</code> fits our pattern with <code>X = m</code>, <code>e1 = (λn. (λa. (λb. (m ((n a) b)) b))))</code> and <code>e2 = (λf. (λx. x))</code>. So after substitution we get <code>(λn. (λa. (λb. ((λf. (λx. x)) ((n a) b)) b)))</code>, which makes our whole expression:</p> <pre><code>(λn. (λa. (λb. ((λf. (λx. x)) ((n a) b)) b))) (λf. (λx. (f x))) </code></pre> <p>Now we can apply the pattern to the whole expression with <code>X = n</code>, <code>e1 = (λa. (λb. ((λf. (λx. x)) ((n a) b)) b))</code> and <code>e2 = (λf. (λx. (f x)))</code>. So after substituting we get:</p> <pre><code>(λa. (λb. ((λf. (λx. x)) (((λf. (λx. (f x))) a) b)) b)) </code></pre> <p>Now <code>((λf. (λx. (f x))) a)</code> fits our pattern and becomes <code>(λx. (a x))</code>, which leads to:</p> <pre><code>(λa. (λb. ((λf. (λx. x)) ((λx. (a x)) b)) b)) </code></pre> <p>This time we can apply the pattern to <code>((λx. (a x)) b)</code>, which reduces to <code>(a b)</code>, leading to:</p> <pre><code>(λa. (λb. ((λf. (λx. x)) (a b)) b)) </code></pre> <p>Now apply the pattern to <code>((λf. (λx. x)) (a b))</code>, which reduces to <code>(λx. x)</code> and get:</p> <pre><code>(λa. (λb. b)) </code></pre> <p>Now we're done.</p>
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    1. COAwesome ! Thanks sepp2k
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    2. COsepp2k: I have a question, should the reduction be done from left to right or the other way round? Or does it not matter?
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    3. COYou can't get different answers by reducing in a different order. However, doing it one way might keep reducing forever. To avoid this you can use "normal-order reduction" which reduces the left most first. (Or, more precisely left most and outer most - basically the one that starts the furthest to the left.) This is guaranteed to give an answer if one exists.
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