By Harper R.

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**Example text**

Xk ←b1 , . . , bk ] xi = bi abt (1 ≤ i ≤ n ) x1 # y · · · x k # y [ x1 , . . , xk ←b1 , . . , bk ]y = y abt [ x1 , . . , xk ←b1 , . . , bk ] β 1 = β 1 abs · · · [ x1 , . . , xk ←b1 , . . , bk ] β n = β n abs [ x1 , . . , xk ←b1 , . . , bk ]o ( β 1 , . . , β n ) = o ( β 1 , . . , β n ) abt [ x1 , . . , xk ←b1 , . . , bk ] a = a abt [ x1 , . . , xk ←b1 , . . , bk ] a = a abs y # b1 · · · y # bk [ x1 , . . , xk ←b1 , . . , bk ] β = β abs [ x1 , . . , xk ←b1 , . . β abs The apartness conditions on the last clause can always be met, by the freshness assumption.

The most interesting case, of course, is the last. For the sake of concision, let us assume that the abstractor binds one variable, x; the general case is the same, only a bit more awkward to state. The inductive hypothesis states that if d(b abt ) = 0, then there exists a unique n such that d([ x1 , . . , xk , x ← a1 , . . , ak , b] a abt ) = n. [ x1 , . . , xk ← a1 , . . , ak ]b abs ) = n . The result follows immediately by taking, in the inductive hypothesis, b = x, and taking, in the conclusion, n = n + 1.

Xk ← a1 , . . , ak ] a abt ) = n. 2. If d( x1 abt ) = 0, . . ,xk d( β abs ) = n, then for every family a1 , . . , ak of abt’s such that d( a1 abt ) = 0, . . , d( ak abt ) = 0, there exists a unique n such that d([ x1 , . . , xk ← a1 , . . , ak ] β abt ) = n. We may prove these facts simultaneously by rule induction. The most interesting case, of course, is the last. For the sake of concision, let us assume that the abstractor binds one variable, x; the general case is the same, only a bit more awkward to state.