The process a or process b enters the critical section, but not at the same time. If this happens (i.e. they enter the critical section simultenously), an interrupt will be executed.
p = process a
q = process b
r = critical section
operator ∨ = or
operator → = implication
my answer :
(p ∨ q) → r
Is this correct or have I Done something wrong? I am trying to understand propositional logic.
Related
Logical negation (NOT), conjunction (AND), disjunction (OR) and exclusive disjunction (XOR) are very often used in every programming language. Why isn't the material implication (p → q) used so often?
I'd say the ubiquitous "if statement" represents material implication
if p then q
If p is true then q follows: p->q
So I'm trying to define a function apply_C :: "('a multiset ⇒ 'a option) ⇒ 'a multiset ⇒ 'a multiset"
It takes in a function C that may convert an 'a multiset into a single element of type 'a. Here we assume that each element in the domain of C is pairwise mutually exclusive and not the empty multiset (I already have another function that checks these things). apply will also take another multiset inp. What I'd like the function to do is check if there is at least one element in the domain of C that is completely contained in inp. If this is the case, then perform a set difference inp - s where s is the element in the domain of C and add the element the (C s) into this resulting multiset. Afterwards, keep running the function until there are no more elements in the domain of C that are completely contained in the given inp multiset.
What I tried was the following:
fun apply_C :: "('a multiset ⇒ 'a option) ⇒ 'a multiset ⇒ 'a multiset" where
"apply_C C inp = (if ∃s ∈ (domain C). s ⊆# inp then apply_C C (add_mset (the (C s)) (inp - s)) else inp)"
However, I get this error:
Variable "s" occurs on right hand side only:
⋀C inp s.
apply_C C inp =
(if ∃s∈domain C. s ⊆# inp
then apply_C C
(add_mset (the (C s)) (inp - s))
else inp)
I have been thinking about this problem for days now, and I haven't been able to find a way to implement this functionality in Isabelle. Could I please have some help?
After thinking more about it, I don't believe there is a simple solutions for that Isabelle.
Do you need that?
I have not said why you want that. Maybe you can reduce your assumptions? Do you really need a function to calculate the result?
How to express the definition?
I would use an inductive predicate that express one step of rewriting and prove that the solution is unique. Something along:
context
fixes C :: ‹'a multiset ⇒ 'a option›
begin
inductive apply_CI where
‹apply_CI (M + M') (add_mset (the (C M)) M')›
if ‹M ∈ dom C›
context
assumes
distinct: ‹⋀a b. a ∈ dom C ⟹ b ∈ dom C ⟹ a ≠ b ⟹ a ∩# b = {#}› and
strictly_smaller: ‹⋀a b. a ∈ dom C ⟹ size a > 1›
begin
lemma apply_CI_determ:
assumes
‹apply_CI⇧*⇧* M M⇩1› and
‹apply_CI⇧*⇧* M M⇩2› and
‹⋀M⇩3. ¬apply_CI M⇩1 M⇩3›
‹⋀M⇩3. ¬apply_CI M⇩2 M⇩3›
shows ‹M⇩1 = M⇩2›
sorry
lemma apply_CI_smaller:
‹apply_CI M M' ⟹ size M' ≤ size M›
apply (induction rule: apply_CI.induct)
subgoal for M M'
using strictly_smaller[of M]
by auto
done
lemma wf_apply_CI:
‹wf {(x, y). apply_CI y x}›
(*trivial but very annoying because not enough useful lemmas on wf*)
sorry
end
end
I have no clue how to prove apply_CI_determ (no idea if the conditions I wrote down are sufficient or not), but I did spend much thinking about it.
After that you can define your definitions with:
definition apply_C where
‹apply_C M = (SOME M'. apply_CI⇧*⇧* M M' ∧ (∀M⇩3. ¬apply_CI M' M⇩3))›
and prove the property in your definition.
How to execute it
I don't see how to write an executable function on multisets directly. The problem you face is that one step of apply_C is nondeterministic.
If you can use lists instead of multisets, you get an order on the elements for free and you can use subseqs that gives you all possible subsets. Rewrite using the first element in subseqs that is in the domain of C. Iterate as long as there is any possible rewriting.
Link that to the inductive predicate to prove termination and that it calculates the right thing.
Remark that in general you cannot extract a list out of a multiset, but it is possible to do so in some cases (e.g., if you have a linorder over 'a).
How can I tell Isabelle to expand all my definitions, please, because that way the proof is trivial? Unfortunately there is no default expansion or simplification happens, and basically I get back the original expression as the subgoal.
Example:
theory Test
imports Main
begin
definition b0 :: "nat⇒nat"
where "b0 n ≡ (n mod 2)"
definition b1 :: "nat⇒nat"
where "b1 n ≡ (n div 2)"
lemma "(a::nat)≤3 ∧ (b::nat)≤3 ⟶
2*(b1 a)+(b0 a)+2*(b1 b)+(b0 b) = a+b"
apply auto
oops
end
Respose before oops:
proof (prove)
goal (1 subgoal):
1. a ≤ 3 ⟹
b ≤ 3 ⟹ 2 * b1 a + b0 a + 2 * b1 b + b0 b = a + b
My recommendation: unfolding
There is a special keyword unfolding for unpacking definitions at the start of proofs. For your example this would read:
unfolding b0_def b1_def by simp
I consider unfolding the most elegant way. It also helps while writing the proofs. Internally, this is (mostly?) equivalent to using the unfold-method:
apply (unfold b0_def b1_def) by simp
This will recursively (!) use the set of equalities you supply to rewrite the proof goal. (Due to the recursion, you should rather not supply a set of equalities that could generate cycles...)
Alternative: Using the simplifier
In cases with possible loops, the simplifier might be able to reach a nice unfolding without running into these cycles, maybe by interleaving with other simplifications. In such cases, by (simp add: b0_def b1_def), as you've suggested, is great!
Alternative definition: Maybe it's just an abbreviation (and no definition)?
If you find yourself unfolding a definition in every single instance, you could consider, using abbreviation instead of definition. Then, some Isabelle magic will do the packing/unpacking for you without further hints. abbeviation does only affect how the user communicates with Isabelle. It does not introduce new symbols at the object level, and consequently, there would be no b1_def facts and the like.
abbreviation b0 :: "nat⇒nat"
where "b0 n ≡ (n mod 2)"
Usually not recommended: Building something like an abbreviation using the simplifier
If you (for whatever reason) want to have a defined name at the object level, but unfold it in almost every instance, you can also feed the defining equality directly into the simplifier.
definition b0 :: "nat⇒nat"
where [simp]: "b0 n ≡ (n mod 2)"
(Usually there should be little reason for the last option.)
Yes, I keep forgetting that definitions are not used in simplifications by default.
Adding the definitions explicitly to the simplification rules solves this problem:
lemma "(a::nat)≤3 ∧ (b::nat)≤3 ⟶
2*(b1 a)+(b0 a)+2*(b1 b)+(b0 b) = a+b"
by (simp add: b0_def b1_def)
This way the definitions (b0, b1) are correctly used.
When I have a goal such as "∀x. P x" in Isabelle, I know that I can write
show "∀x. P x"
proof (rule allI)
However, when the goal is "∀x>0. P x", I cannot do that. Is there a similar rule/method that I can use after proof in order to simplify my goal? I would also be interested in one for the situation where you have a goal of the form "∃x>0. P x".
I'm looking for an Isar proof that uses the proof (rule something) style.
Universal quantifier
To expand on Lars's answer: ∀x>0. P x is just syntactic sugar for ∀x. x > 0 ⟶ P x. As a consequence, if you want to prove a statement like this, you first have to strip away the universal quantifier with allI and then strip away the implication with impI. You can do something like this:
lemma "∀x>0. P x"
proof (rule allI, rule impI)
Or using intro, which is more or less the same as applying rule until it is not possible anymore:
lemma "∀x>0. P x"
proof (intro allI impI)
Or you can use safe, which eagerly applies all introduction rules that are declared as ‘safe’, such as allI and impI:
lemma "∀x>0. P x"
proof safe
In any case, your new proof state is then
proof (state)
goal (1 subgoal):
1. ⋀x. 0 < x ⟹ P x
And you can proceed like this:
lemma "∀x>0. P (x :: nat)"
proof safe
fix x :: nat assume "x > 0"
show "P x"
Note that I added an annotation; I didn't know what type your P has, so I just used nat. When you fix a variable in Isar and the type is not clear from the assumptions, you will get a warning that a new free type variable was introduced, which is not what you want. When you get that warning, you should add a type annotation to the fix like I did above.
Existential quantifier
For an existential quantifier, safe will not work because the intro rule exI is not always safe due to technical reasons. The typical proof pattern for an ∃x>0. P x would be something like:
lemma "∃x>0. P (x :: nat)"
proof -
have "42 > (0 :: nat)" by simp
moreover have "P 42" sorry
ultimately show ?thesis by blast
qed
Or a little more explicitly:
lemma "∃x>0. P (x :: nat)"
proof -
have "42 > 0 ∧ P 42" sorry
thus ?thesis by (rule exI)
qed
In cases when the existential witness (i.e. the 42 in this example) does not depend on any variables that you got out of an obtain command, you can also do it more directly:
lemma "∃x>0. P (x :: nat)"
proof (intro exI conjI)
This leaves you with the goals ?x > 0 and P ?x. Note that the ?x is a schematic variable for which you can put it anything. So you can complete the proof like this:
lemma "∃x>0. P (x :: nat)"
proof (intro exI conjI)
show "42 > (0::nat)" by simp
show "P 42" sorry
qed
As I said, this does not work if your existential witness depends on some variable that you got from obtain due to technical restrictions. In that case, you have to fall back to the other solution I mentioned.
The following works in Isabelle2016-1-RC2:
lemma "∀ x>0. P x"
apply (rule allI)
In general, you can also just use apply rule, which will select the default introduction rule. Same is true for the existential quantifier.
Click here for the answer. Turing Machine
The question is to construct a Turing Machine which accepts the regular expression,
L = {a^n b^n | n>= 1}.
I am not sure if my answer is correct or wrong. Thank you in advance for your reply.
You cannot "accept the regular expression", only the language it describes. And what you provide is not a regular expression, but a set description. In fact, the language is not regular and therefore cannot be described by standard regular expressions.
The machine from your answer accepts the language described by the regular expression a^+ b^+.
A TM could mark the first a (e.g. by converting it to A) then delete the first b. And for each n one loop. If you and up with a string only of A, then accept.
As stated before, language L = {a^nb^n; n >= 1} cannot be described by regular expressions, it doesn't belong into the category of regular grammars. This language in particular is an example of context-free grammar, and thus it can be described by context-free grammar and recognized by pushdown automaton (an automaton with LIFO memory, a stack).
Grammar for this language would look something like this:
G = (V, S, R, P)
Where:
V is finite set of non-terminal characters, V = { S }
S is finite set of terminal characters, S = { a, b }
R is relation that describes "rewrites" from non-terminal characters to non-terminals and terminals, in this case R = { S -> aSb, S -> ab }
P is starting non-terminal character, P = S
A pushdown automata recognizing this language would be more complex, as it is a 7-tuple M = (Q, S, G, D, q0, Z, F)
Q is set of states
S is input alphabet
G is stack alphabet
D is the transition relation
q0 is start state
Z is initial stack symbol
F is set of accepting states
For our case, it would be:
Q = { q0, q1, qF }
S = { a, b }
G = { z0, X }
D will take a form of relation (current state, input character, top of stack) -> (output state, top of stack) (meaning you can move to a different state and rewrite top of stack (erase it, rewrite it or let it be)
(q0, a, z0) -> (q0, Xz0) - reading the first a
(q0, a, X) -> (q0, XX) - reading consecutive a's
(q0, b, X) -> (q1, e) - reading first b
(q1, b, X) -> (q1, e) - reading consecutive b's
(q1, e, z0) -> (qF, e) - reading last b
where e is empty word (sometimes called epsilon)
q0 = q0
Z = z0
F = { qF }
The language L = {a^n b^n | n≥1} represents a kind of language where we use only 2 character, i.e., a, b. In the beginning language has some number of a’s followed by equal number of b’s . Any such string which falls in this category will be accepted by this language. The beginning and end of string is marked by $ sign.
Step-1:
Replace a by X and move right, Go to state Q1.
Step-2:
Replace a by a and move right, Remain on same state
Replace Y by Y and move right, Remain on same state
Replace b by Y and move right, go to state Q2.
Step-3:
Replace b by b and move left, Remain on same state
Replace a by a and move left, Remain on same state
Replace Y by Y and move left, Remain on same state
Replace X by X and move right, go to state Q0.
Step-5:
If symbol is Y replace it by Y and move right and Go to state Q4
Else go to step 1
Step-6:
Replace Y by Y and move right, Remain on same state
If symbol is $ replace it by $ and move left, STRING IS ACCEPTED, GO TO FINAL STATE Q4