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Up to ((λz.w) (λz.w)), everything is fine. Now (λz.w) is the argument to be substituted for 0 occurrences of z in w; so the result of substitution will be w itself.
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Lambda Calculus - Reductions #2
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Up to ((λz.w) (λz.w)), everything is fine. Now (λz.w) is the argument to be substituted for 0 occurrences of z in w; so the result of substitution will be w itself.