Quantifier-free way to specify constructor of Z3 sum typeWhat methods does Z3 use to solve quantifier-free bit-vector formulas (QF_BV)?Equivalent Quantifier Free FormulasWhat is the correct way to handle quantified formulas with respect to empty models?Quantifier elimination for enumeration types in Z3Is z3 the most efficient solver for quantifier-free integer propositional logic?Is there a way to maximize sum using minimum entries from a list?What's the decision procedure for quantifier-free recursive lists in z3?z3 datatype matching without a quantifierWhat does “quantifier free logic” mean in SMT context?
Is it possible to have the age of the universe be unknown?
Extreme flexible working hours: how to control people and activities?
What setting controls moving the cursor on the command line?
How is water heavier than petrol, even though its molecular weight is less than petrol?
What speaks against investing in precious metals?
SQL counting distinct over partition
Geopandas and QGIS Calulating Different Polygon Area Values?
How can this tool find out registered domains from an IP?
Teaching a class likely meant to inflate the GPA of student athletes
Is using haveibeenpwned to validate password strength rational?
Are there any important biographies of nobodies?
How did old MS-DOS games utilize various graphic cards?
I have a problem assistant manager, but I can't fire him
How to use memset in c++?
Rebus with 20 song titles
Switch "when" cannot see constants?
Is it expected that a reader will skip parts of what you write?
Why we don’t make use of the t-distribution for constructing a confidence interval for a proportion?
Check if three arrays contains the same element
Why can my keyboard only digest 6 keypresses at a time?
How do I prevent employees from either switching to competitors or opening their own business?
Bent Peugeot Carbolite 103 Frame
What to do when surprise and a high initiative roll conflict with the narrative?
When would it be advantageous not apply Training Ground's cost reduction?
Quantifier-free way to specify constructor of Z3 sum type
What methods does Z3 use to solve quantifier-free bit-vector formulas (QF_BV)?Equivalent Quantifier Free FormulasWhat is the correct way to handle quantified formulas with respect to empty models?Quantifier elimination for enumeration types in Z3Is z3 the most efficient solver for quantifier-free integer propositional logic?Is there a way to maximize sum using minimum entries from a list?What's the decision procedure for quantifier-free recursive lists in z3?z3 datatype matching without a quantifierWhat does “quantifier free logic” mean in SMT context?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty height:90px;width:728px;box-sizing:border-box;
Let's say I have a simple sum type in Z3 with several constructors of different arities:
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
How can I assert that I know value of type Foo
is a quux
? I could introduce an existential over quux1
and quux2
but I'm wary of introducing seemingly unnecessary quantifiers. Is there a better way to assert this?
z3 smt
add a comment |
Let's say I have a simple sum type in Z3 with several constructors of different arities:
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
How can I assert that I know value of type Foo
is a quux
? I could introduce an existential over quux1
and quux2
but I'm wary of introducing seemingly unnecessary quantifiers. Is there a better way to assert this?
z3 smt
add a comment |
Let's say I have a simple sum type in Z3 with several constructors of different arities:
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
How can I assert that I know value of type Foo
is a quux
? I could introduce an existential over quux1
and quux2
but I'm wary of introducing seemingly unnecessary quantifiers. Is there a better way to assert this?
z3 smt
Let's say I have a simple sum type in Z3 with several constructors of different arities:
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
How can I assert that I know value of type Foo
is a quux
? I could introduce an existential over quux1
and quux2
but I'm wary of introducing seemingly unnecessary quantifiers. Is there a better way to assert this?
z3 smt
z3 smt
asked Mar 24 at 17:52
copumpkincopumpkin
2,2921321
2,2921321
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Existentials are rather harmless when assumed because the existentially quantified variables are directly instantiated and thus only entail additional symbols. That is, the second half of the snippet
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
(declare-const f Foo)
(assert (exists ((s String) (i Int)) ;; Let f be a quux
(= f (quux s i))
))
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
is equivalent to
...
(declare-const _s String)
(declare-const _i Int)
(assert (= f (quux _s _i))) ;; Let f be a quux
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
If you're wary of existentials but not of foralls, then you could tag Foo
values by axiomatising a mapping from Foo
constructors to distinct tags:
(set-option :smt.mbqi false)
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
;; Declare a finite sort Foo_tag with three distinct elements
(declare-datatypes () ((Foo_tag Foo_tag.bar Foo_tag.baz Foo_tag.quux)))
;; Alternatively, three distinct elements from an infinite sort such
;; as Int can be used. Either by choosing distinct but unspecified
;; values, as done below, or by directly choosing concrete values,
;; e.g. 1, 2, 3.
; (define-sort Foo_tag () Int)
; (declare-const Foo_tag.bar Foo_tag)
; (declare-const Foo_tag.baz Foo_tag)
; (declare-const Foo_tag.quux Foo_tag)
; (assert (distinct Foo_tag.bar Foo_tag.baz Foo_tag.quux))
;; Tagging function
(declare-fun tag_of (Foo) Foo_tag)
;; Tagging axiom for bar ...
(assert (= (tag_of bar) Foo_tag.bar))
;; ... baz ...
(assert (forall ((s String)) (!
(= (tag_of (baz s)) Foo_tag.baz)
:pattern ((baz s))
)))
;; ... and quux
(assert (forall ((s String) (i Int)) (!
(= (tag_of (quux s i)) Foo_tag.quux)
:pattern ((quux s i))
)))
;; Let's do some testing
(declare-const f Foo)
(assert (= (tag_of f) Foo_tag.quux)) ;; Tag f as a quux
(push)
(assert (= f bar))
(check-sat) ;; UNSAT - as expected
(pop)
(push)
(assert (= f (baz "test")))
(check-sat) ;; UNSAT - as expected
(pop)
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "1"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55326751%2fquantifier-free-way-to-specify-constructor-of-z3-sum-type%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Existentials are rather harmless when assumed because the existentially quantified variables are directly instantiated and thus only entail additional symbols. That is, the second half of the snippet
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
(declare-const f Foo)
(assert (exists ((s String) (i Int)) ;; Let f be a quux
(= f (quux s i))
))
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
is equivalent to
...
(declare-const _s String)
(declare-const _i Int)
(assert (= f (quux _s _i))) ;; Let f be a quux
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
If you're wary of existentials but not of foralls, then you could tag Foo
values by axiomatising a mapping from Foo
constructors to distinct tags:
(set-option :smt.mbqi false)
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
;; Declare a finite sort Foo_tag with three distinct elements
(declare-datatypes () ((Foo_tag Foo_tag.bar Foo_tag.baz Foo_tag.quux)))
;; Alternatively, three distinct elements from an infinite sort such
;; as Int can be used. Either by choosing distinct but unspecified
;; values, as done below, or by directly choosing concrete values,
;; e.g. 1, 2, 3.
; (define-sort Foo_tag () Int)
; (declare-const Foo_tag.bar Foo_tag)
; (declare-const Foo_tag.baz Foo_tag)
; (declare-const Foo_tag.quux Foo_tag)
; (assert (distinct Foo_tag.bar Foo_tag.baz Foo_tag.quux))
;; Tagging function
(declare-fun tag_of (Foo) Foo_tag)
;; Tagging axiom for bar ...
(assert (= (tag_of bar) Foo_tag.bar))
;; ... baz ...
(assert (forall ((s String)) (!
(= (tag_of (baz s)) Foo_tag.baz)
:pattern ((baz s))
)))
;; ... and quux
(assert (forall ((s String) (i Int)) (!
(= (tag_of (quux s i)) Foo_tag.quux)
:pattern ((quux s i))
)))
;; Let's do some testing
(declare-const f Foo)
(assert (= (tag_of f) Foo_tag.quux)) ;; Tag f as a quux
(push)
(assert (= f bar))
(check-sat) ;; UNSAT - as expected
(pop)
(push)
(assert (= f (baz "test")))
(check-sat) ;; UNSAT - as expected
(pop)
add a comment |
Existentials are rather harmless when assumed because the existentially quantified variables are directly instantiated and thus only entail additional symbols. That is, the second half of the snippet
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
(declare-const f Foo)
(assert (exists ((s String) (i Int)) ;; Let f be a quux
(= f (quux s i))
))
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
is equivalent to
...
(declare-const _s String)
(declare-const _i Int)
(assert (= f (quux _s _i))) ;; Let f be a quux
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
If you're wary of existentials but not of foralls, then you could tag Foo
values by axiomatising a mapping from Foo
constructors to distinct tags:
(set-option :smt.mbqi false)
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
;; Declare a finite sort Foo_tag with three distinct elements
(declare-datatypes () ((Foo_tag Foo_tag.bar Foo_tag.baz Foo_tag.quux)))
;; Alternatively, three distinct elements from an infinite sort such
;; as Int can be used. Either by choosing distinct but unspecified
;; values, as done below, or by directly choosing concrete values,
;; e.g. 1, 2, 3.
; (define-sort Foo_tag () Int)
; (declare-const Foo_tag.bar Foo_tag)
; (declare-const Foo_tag.baz Foo_tag)
; (declare-const Foo_tag.quux Foo_tag)
; (assert (distinct Foo_tag.bar Foo_tag.baz Foo_tag.quux))
;; Tagging function
(declare-fun tag_of (Foo) Foo_tag)
;; Tagging axiom for bar ...
(assert (= (tag_of bar) Foo_tag.bar))
;; ... baz ...
(assert (forall ((s String)) (!
(= (tag_of (baz s)) Foo_tag.baz)
:pattern ((baz s))
)))
;; ... and quux
(assert (forall ((s String) (i Int)) (!
(= (tag_of (quux s i)) Foo_tag.quux)
:pattern ((quux s i))
)))
;; Let's do some testing
(declare-const f Foo)
(assert (= (tag_of f) Foo_tag.quux)) ;; Tag f as a quux
(push)
(assert (= f bar))
(check-sat) ;; UNSAT - as expected
(pop)
(push)
(assert (= f (baz "test")))
(check-sat) ;; UNSAT - as expected
(pop)
add a comment |
Existentials are rather harmless when assumed because the existentially quantified variables are directly instantiated and thus only entail additional symbols. That is, the second half of the snippet
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
(declare-const f Foo)
(assert (exists ((s String) (i Int)) ;; Let f be a quux
(= f (quux s i))
))
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
is equivalent to
...
(declare-const _s String)
(declare-const _i Int)
(assert (= f (quux _s _i))) ;; Let f be a quux
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
If you're wary of existentials but not of foralls, then you could tag Foo
values by axiomatising a mapping from Foo
constructors to distinct tags:
(set-option :smt.mbqi false)
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
;; Declare a finite sort Foo_tag with three distinct elements
(declare-datatypes () ((Foo_tag Foo_tag.bar Foo_tag.baz Foo_tag.quux)))
;; Alternatively, three distinct elements from an infinite sort such
;; as Int can be used. Either by choosing distinct but unspecified
;; values, as done below, or by directly choosing concrete values,
;; e.g. 1, 2, 3.
; (define-sort Foo_tag () Int)
; (declare-const Foo_tag.bar Foo_tag)
; (declare-const Foo_tag.baz Foo_tag)
; (declare-const Foo_tag.quux Foo_tag)
; (assert (distinct Foo_tag.bar Foo_tag.baz Foo_tag.quux))
;; Tagging function
(declare-fun tag_of (Foo) Foo_tag)
;; Tagging axiom for bar ...
(assert (= (tag_of bar) Foo_tag.bar))
;; ... baz ...
(assert (forall ((s String)) (!
(= (tag_of (baz s)) Foo_tag.baz)
:pattern ((baz s))
)))
;; ... and quux
(assert (forall ((s String) (i Int)) (!
(= (tag_of (quux s i)) Foo_tag.quux)
:pattern ((quux s i))
)))
;; Let's do some testing
(declare-const f Foo)
(assert (= (tag_of f) Foo_tag.quux)) ;; Tag f as a quux
(push)
(assert (= f bar))
(check-sat) ;; UNSAT - as expected
(pop)
(push)
(assert (= f (baz "test")))
(check-sat) ;; UNSAT - as expected
(pop)
Existentials are rather harmless when assumed because the existentially quantified variables are directly instantiated and thus only entail additional symbols. That is, the second half of the snippet
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
(declare-const f Foo)
(assert (exists ((s String) (i Int)) ;; Let f be a quux
(= f (quux s i))
))
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
is equivalent to
...
(declare-const _s String)
(declare-const _i Int)
(assert (= f (quux _s _i))) ;; Let f be a quux
(assert (= f (baz "test"))) ;; Also let f be a baz
(check-sat) ;; UNSAT - as expected
If you're wary of existentials but not of foralls, then you could tag Foo
values by axiomatising a mapping from Foo
constructors to distinct tags:
(set-option :smt.mbqi false)
(declare-datatypes ()
((Foo bar
(baz (unbaz String))
(quux (unquux1 String) (unquux2 Int)))))
;; Declare a finite sort Foo_tag with three distinct elements
(declare-datatypes () ((Foo_tag Foo_tag.bar Foo_tag.baz Foo_tag.quux)))
;; Alternatively, three distinct elements from an infinite sort such
;; as Int can be used. Either by choosing distinct but unspecified
;; values, as done below, or by directly choosing concrete values,
;; e.g. 1, 2, 3.
; (define-sort Foo_tag () Int)
; (declare-const Foo_tag.bar Foo_tag)
; (declare-const Foo_tag.baz Foo_tag)
; (declare-const Foo_tag.quux Foo_tag)
; (assert (distinct Foo_tag.bar Foo_tag.baz Foo_tag.quux))
;; Tagging function
(declare-fun tag_of (Foo) Foo_tag)
;; Tagging axiom for bar ...
(assert (= (tag_of bar) Foo_tag.bar))
;; ... baz ...
(assert (forall ((s String)) (!
(= (tag_of (baz s)) Foo_tag.baz)
:pattern ((baz s))
)))
;; ... and quux
(assert (forall ((s String) (i Int)) (!
(= (tag_of (quux s i)) Foo_tag.quux)
:pattern ((quux s i))
)))
;; Let's do some testing
(declare-const f Foo)
(assert (= (tag_of f) Foo_tag.quux)) ;; Tag f as a quux
(push)
(assert (= f bar))
(check-sat) ;; UNSAT - as expected
(pop)
(push)
(assert (= f (baz "test")))
(check-sat) ;; UNSAT - as expected
(pop)
answered Apr 24 at 10:22
Malte SchwerhoffMalte Schwerhoff
9,49043061
9,49043061
add a comment |
add a comment |
Thanks for contributing an answer to Stack Overflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstackoverflow.com%2fquestions%2f55326751%2fquantifier-free-way-to-specify-constructor-of-z3-sum-type%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown