There are two redexes in this term; we've marked the operative lambdas
with a star. If we reduce the leftmost redex first, the term reduces
There are two redexes in this term; we've marked the operative lambdas
with a star. If we reduce the leftmost redex first, the term reduces
redex instead, the "reduced" form is `(\x.I)Ω` again, and we are in
danger of entering an infinite loop.
redex instead, the "reduced" form is `(\x.I)Ω` again, and we are in
danger of entering an infinite loop.