Instructions Prove the reduction below using Natural Semantics. This assignment does not require the use of side conditions. To use the tool below: • [+] adds a subproof above the line to the selected inference. • [e] allows you to edit the selected inference. • • [x] deletes the selected inference and its subtrees. [sc] allows you to add a side condition to the selected inference. You may find the natural semantics rules by going here. Please note that every inference is labeled with the rule of which it is an instance. Use the drop-down menu in the tool to select the appropriate label. When you click on [+], you will be given two boxes: the left is the term to be reduced, and the right is what it reduces to. You should enter only one reduction per click of [+]. If your proof requires more than one antecedent, you must repeatedly click on the [+] on the line of the inference to add more antecedents. Some rules require that a relation such as U②V=b must hold as well. You need not add an antecedent proving this, however the condition will still be checked internally. Problem [+] | [sc] Num [e] [x] Num (2, {x -> 0, y > 0}) Ident الله 2 [+] | [sc] [e] [x] (1, {x -> 0, y > 0}) (x, {x -> 2, y > 0}) th 2 1 Assign (x = 2, {x > 0, y > 0}) -> {x -> 2, y > 0} Rel While Seq [+] | [sc] [e] | [x] [+] | [sc] [e] | [x] [+] [se] [e] | [x] (y = 1, {x > 0, y => 0}) -> false (while y 1 do skip od, {x -> 0, y -> 0}) {x -> 0, y => 0} (while y 1 do skip od; x = 2, {x -> 0, y => 0}) ⇓ {x -> 2, y => 0} [+] | [sc] [e] [x] [+] | [sc]
Instructions Prove the reduction below using Natural Semantics. This assignment does not require the use of side conditions. To use the tool below: • [+] adds a subproof above the line to the selected inference. • [e] allows you to edit the selected inference. • • [x] deletes the selected inference and its subtrees. [sc] allows you to add a side condition to the selected inference. You may find the natural semantics rules by going here. Please note that every inference is labeled with the rule of which it is an instance. Use the drop-down menu in the tool to select the appropriate label. When you click on [+], you will be given two boxes: the left is the term to be reduced, and the right is what it reduces to. You should enter only one reduction per click of [+]. If your proof requires more than one antecedent, you must repeatedly click on the [+] on the line of the inference to add more antecedents. Some rules require that a relation such as U②V=b must hold as well. You need not add an antecedent proving this, however the condition will still be checked internally. Problem [+] | [sc] Num [e] [x] Num (2, {x -> 0, y > 0}) Ident الله 2 [+] | [sc] [e] [x] (1, {x -> 0, y > 0}) (x, {x -> 2, y > 0}) th 2 1 Assign (x = 2, {x > 0, y > 0}) -> {x -> 2, y > 0} Rel While Seq [+] | [sc] [e] | [x] [+] | [sc] [e] | [x] [+] [se] [e] | [x] (y = 1, {x > 0, y => 0}) -> false (while y 1 do skip od, {x -> 0, y -> 0}) {x -> 0, y => 0} (while y 1 do skip od; x = 2, {x -> 0, y => 0}) ⇓ {x -> 2, y => 0} [+] | [sc] [e] [x] [+] | [sc]
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
Related questions
Question
Computer Science Natural Semantics
![Instructions
Prove the reduction below using Natural Semantics. This assignment does not require the use of side
conditions.
To use the tool below:
•
[+] adds a subproof above the line to the selected inference.
•
[e] allows you to edit the selected inference.
•
•
[x] deletes the selected inference and its subtrees.
[sc] allows you to add a side condition to the selected inference.
You may find the natural semantics rules by going here.
Please note that every inference is labeled with the rule of which it is an instance. Use the drop-down
menu in the tool to select the appropriate label. When you click on [+], you will be given two boxes: the
left is the term to be reduced, and the right is what it reduces to. You should enter only one reduction
per click of [+]. If your proof requires more than one antecedent, you must repeatedly click on the [+] on
the line of the inference to add more antecedents. Some rules require that a relation such as U②V=b
must hold as well. You need not add an antecedent proving this, however the condition will still be
checked internally.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F840ac9d4-eb27-4280-b70d-21eed9a81895%2F46b3e12f-ec4a-4c47-a226-909d1a044276%2Fmhki7yn_processed.png&w=3840&q=75)
Transcribed Image Text:Instructions
Prove the reduction below using Natural Semantics. This assignment does not require the use of side
conditions.
To use the tool below:
•
[+] adds a subproof above the line to the selected inference.
•
[e] allows you to edit the selected inference.
•
•
[x] deletes the selected inference and its subtrees.
[sc] allows you to add a side condition to the selected inference.
You may find the natural semantics rules by going here.
Please note that every inference is labeled with the rule of which it is an instance. Use the drop-down
menu in the tool to select the appropriate label. When you click on [+], you will be given two boxes: the
left is the term to be reduced, and the right is what it reduces to. You should enter only one reduction
per click of [+]. If your proof requires more than one antecedent, you must repeatedly click on the [+] on
the line of the inference to add more antecedents. Some rules require that a relation such as U②V=b
must hold as well. You need not add an antecedent proving this, however the condition will still be
checked internally.
![Problem
[+] | [sc]
Num
[e] [x]
Num
(2, {x
-> 0, y > 0})
Ident
الله
2
[+] | [sc]
[e] [x]
(1, {x -> 0, y > 0})
(x, {x -> 2, y > 0})
th
2
1
Assign
(x = 2, {x > 0, y > 0})
->
{x -> 2, y > 0}
Rel
While
Seq
[+] | [sc]
[e] | [x]
[+] | [sc]
[e] | [x]
[+] [se]
[e] | [x]
(y = 1, {x > 0, y => 0})
->
false
(while y 1 do skip od, {x -> 0, y -> 0})
{x
-> 0, y => 0}
(while y 1 do skip od; x = 2, {x -> 0, y => 0})
⇓ {x -> 2, y => 0}
[+] | [sc]
[e] [x]
[+] | [sc]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F840ac9d4-eb27-4280-b70d-21eed9a81895%2F46b3e12f-ec4a-4c47-a226-909d1a044276%2Fv11mmfi_processed.png&w=3840&q=75)
Transcribed Image Text:Problem
[+] | [sc]
Num
[e] [x]
Num
(2, {x
-> 0, y > 0})
Ident
الله
2
[+] | [sc]
[e] [x]
(1, {x -> 0, y > 0})
(x, {x -> 2, y > 0})
th
2
1
Assign
(x = 2, {x > 0, y > 0})
->
{x -> 2, y > 0}
Rel
While
Seq
[+] | [sc]
[e] | [x]
[+] | [sc]
[e] | [x]
[+] [se]
[e] | [x]
(y = 1, {x > 0, y => 0})
->
false
(while y 1 do skip od, {x -> 0, y -> 0})
{x
-> 0, y => 0}
(while y 1 do skip od; x = 2, {x -> 0, y => 0})
⇓ {x -> 2, y => 0}
[+] | [sc]
[e] [x]
[+] | [sc]
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