\(\lambda\)-calculus II:
substitution rules
2021 James B. Wilson
Colorado State University
Need functions before sets and types
\(\lambda\)-calculus is functions by substitution
Substitution can go wrong, it needs rules.
Consistant Substitution
- One variable at a time (\(\lambda\)-abstraction).
- Rename variables (\(\alpha\)-conversion).
- Substitute into variables (\(\beta\)-reduction).
- Name functions (\(\eta\)-abstraction).
The funny names are historical and while not memorable, they are still used today in logic and informatics.
Anonymous Functions
"\(\lambda\)'s"
Goal: substitute variable in \(M\) consistently.
Rule: one variable at a time, denote it:
\[x\mapsto M\qquad \lambda x.M\]
This is an anonymous-function, also called a "\(\lambda\)".
Vocabulary:
the variable \(x\) is bound in \(x\mapsto M\)
What about all those 2-variable functions?
\((x,y)\mapsto x+y\) can be treated as one by partial evaluation
\[L_x = y\mapsto x+y\]
\[x\mapsto L_x\]
We could do likewise with \(R_y=x\mapsto x+y\).
Called "Currying" and it needs context to know the order.
First draft Rules
Identity: \(I(x)=x\) means \(I(7)=7\) \[[x:=C](x\mapsto x) \quad \rhd\quad C\]
Constant: \(K_3(x)=3\) means \(K_3(7)=3\) \[[x:=C](x\mapsto A)\quad\rhd\quad A\]
Recursion: \(f(x)=x^2+x\) means \(f(7)=7^2+7\) \[[x:=C](x\mapsto AB)\quad\rhd\quad [x:=C]A [x:=C]B\]
Variable Capture Problem
E.g. \(K_c(x)=c\) is constant until we set \(c=x\) when startlingly it become the identity \(K_x(x)=x=I(x)\).
\[[c:=x](c\mapsto (x\mapsto c))\qquad \rhd \qquad x\mapsto x\]
Solution: rename the variable.
Second draft Rules
- (Unchanged) Identity: \[[x:=C](x\mapsto x) \quad \rhd\quad C\]
- (Unchanged) Recursion: \[[x:=C](x\mapsto AB)\quad\rhd\quad [x:=C]A [x:=C]B\]
- Constant substitutions built term by term...
Substituting into Constants
- If \(A\) is atomic but not \(x\) then constant \[[x:=C](x\mapsto A)\qquad\rhd\qquad A\]
- If \(A\) is a \(x\mapsto B\) i.e. \(\lambda x.B\), then \(x\) is a local variable and essentially invisible outside \(A\), so again constant \[[x:=C](x\mapsto (x\mapsto B))\qquad\rhd\qquad x\mapsto B\]
- If \(A\) is a \(y\mapsto B\) i.e. \(\lambda y.B\) and \(x\) is not a variable in \(B\) then still constant \[[x:=C](x\mapsto (y\mapsto B))\qquad\rhd\qquad y\mapsto B\]
Substituting into Constants
...after renaming!
- \([x:=C](x\mapsto (y\mapsto B))\) and \(x\) is a free variable of \(B\) and \(y\) is not a free variable of \(C\) then first substitute \(x\) in \(B\) \[\rhd\qquad y\mapsto [x:=C]B\]
- \([x:=C](x\mapsto (y\mapsto B))\) and \(x\) is a free variable of \(B\) and \(y\) is a free variable of \(C\) then first rename \(x\) then substitute \[\rhd\qquad y\mapsto [z:=C][z:=x]B\]
Further Reading
- Hindley-Seldin Lambda-Calculus and Combinators, Cambridge U. Press, Chapter 1.
Substitution Rules: Curry-Feys
By James Wilson
Substitution Rules: Curry-Feys
Substitution needs rules. Curry-Feys is one such system and knowing how to substitute properly is worth the work.
