beta reduction: order of substitution
Do we always apply our input to the left most term in a lamda expression?
For instance, take the expressions:
$λP λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP λQ[ ∀x P(x)→Q(x)]]$
$λP. λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP [λQ[ ∀x P(x)→Q(x)]]]$
If we apply an input to each expression, would the input in both expressions replace P? But if that's the case, wouldn't that make the two expressions equivalent?
first-order-logic lambda-calculus
asked Nov 20 at 1:22
Matt
23818
add a comment |
Do we always apply our input to the left most term in a lamda expression?
For instance, take the expressions:
$λP λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP λQ[ ∀x P(x)→Q(x)]]$
$λP. λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP [λQ[ ∀x P(x)→Q(x)]]]$
If we apply an input to each expression, would the input in both expressions replace P? But if that's the case, wouldn't that make the two expressions equivalent?
first-order-logic lambda-calculus
asked Nov 20 at 1:22
Matt
23818
I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23
add a comment |
Do we always apply our input to the left most term in a lamda expression?
For instance, take the expressions:
$λP λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP λQ[ ∀x P(x)→Q(x)]]$
$λP. λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP [λQ[ ∀x P(x)→Q(x)]]]$
If we apply an input to each expression, would the input in both expressions replace P? But if that's the case, wouldn't that make the two expressions equivalent?
first-order-logic lambda-calculus
asked Nov 20 at 1:22
Matt
23818
Do we always apply our input to the left most term in a lamda expression?
For instance, take the expressions:
$λP λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP λQ[ ∀x P(x)→Q(x)]]$
$λP. λQ. ∀x P(x)→Q(x)$ which we can rewrite as $[λP [λQ[ ∀x P(x)→Q(x)]]]$
If we apply an input to each expression, would the input in both expressions replace P? But if that's the case, wouldn't that make the two expressions equivalent?
first-order-logic lambda-calculus
first-order-logic lambda-calculus
asked Nov 20 at 1:22
Matt
23818
asked Nov 20 at 1:22
Matt
23818
asked Nov 20 at 1:22
Matt
23818
asked Nov 20 at 1:22
Matt
23818
asked Nov 20 at 1:22
Matt
23818
23818
I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23
add a comment |
I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23
I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23
I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23
add a comment |
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I don't see a difference between the two terms besides the dot, which has no meaning.
– Couchy311
Nov 21 at 7:23