Why does MatchQ[a, r_ /; Head[r] != Plus] evaluate to False?

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3
down vote

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I'm trying to understand. Why is it that:

MatchQ[a, r_ /; Head[r] != Plus]

Evaluates to:

False

? For me, I would think that, because:

Head[a]

Evaluates to:

Symbol

Where a has no value or expression assigned to it, then this:

MatchQ[a, r_ /; Head[r] != Plus]

Should evaluate to True. Could someone point me in the right direction to understanding this better? Thanks!

pattern-matching head

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asked 1 hour ago

Jmeeks29ig

39518

add a comment | 

up vote
3
down vote

favorite

I'm trying to understand. Why is it that:

MatchQ[a, r_ /; Head[r] != Plus]

Evaluates to:

False

? For me, I would think that, because:

Head[a]

Evaluates to:

Symbol

Where a has no value or expression assigned to it, then this:

MatchQ[a, r_ /; Head[r] != Plus]

Should evaluate to True. Could someone point me in the right direction to understanding this better? Thanks!

pattern-matching head

share|improve this question

asked 1 hour ago

Jmeeks29ig

39518

add a comment | 

up vote
3
down vote

favorite

up vote
3
down vote

favorite

I'm trying to understand. Why is it that:

MatchQ[a, r_ /; Head[r] != Plus]

Evaluates to:

False

? For me, I would think that, because:

Head[a]

Evaluates to:

Symbol

Where a has no value or expression assigned to it, then this:

MatchQ[a, r_ /; Head[r] != Plus]

Should evaluate to True. Could someone point me in the right direction to understanding this better? Thanks!

pattern-matching head

share|improve this question

asked 1 hour ago

Jmeeks29ig

39518

I'm trying to understand. Why is it that:

MatchQ[a, r_ /; Head[r] != Plus]

Evaluates to:

False

? For me, I would think that, because:

Head[a]

Evaluates to:

Symbol

Where a has no value or expression assigned to it, then this:

MatchQ[a, r_ /; Head[r] != Plus]

Should evaluate to True. Could someone point me in the right direction to understanding this better? Thanks!

pattern-matching head

pattern-matching head

share|improve this question

asked 1 hour ago

Jmeeks29ig

39518

share|improve this question

asked 1 hour ago

Jmeeks29ig

39518

share|improve this question

share|improve this question

asked 1 hour ago

Jmeeks29ig

39518

asked 1 hour ago

Jmeeks29ig

39518

asked 1 hour ago

Jmeeks29ig

39518

39518

add a comment | 

add a comment | 

1 Answer
1

active

oldest

votes

up vote
5
down vote



accepted

The problem is this:

If a is a symbol then its Head is Symbol, so Head[a] =!= Plus reduces to Symbol != Plus. Unequal (!=) is supposed to be a mathematical test for inequality. In this case, it just cannot decide whether Symbol != Plus should evaluate to True or to False, since both sides are Symbols. Here an example why this is undecidable with the current amount of information:

With[Symbol = 1, Plus = 1, Symbol != Plus]
With[Symbol = 1, Plus = 0, Symbol != Plus]

False

True

So the expression Symbol != Plus stays unevaluated. (This is the best strategy since later definitions could make it decidable.)

Because the second argument of Condition (/;) does not evaluate to True, the pattern does not match.

Lesson to learn: For testing for structual inequality, use UnsameQ (=!=):

MatchQ[a, r_ /; Head[r] =!= Plus]

True

Of course, the same applies, mutatis mutandis, to Equal and SameQ.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, thanks! That helps a lot!
  – Jmeeks29ig
  1 hour ago
  

  
  
  
  

add a comment | 

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
5
down vote



accepted

The problem is this:

If a is a symbol then its Head is Symbol, so Head[a] =!= Plus reduces to Symbol != Plus. Unequal (!=) is supposed to be a mathematical test for inequality. In this case, it just cannot decide whether Symbol != Plus should evaluate to True or to False, since both sides are Symbols. Here an example why this is undecidable with the current amount of information:

With[Symbol = 1, Plus = 1, Symbol != Plus]
With[Symbol = 1, Plus = 0, Symbol != Plus]

False

True

So the expression Symbol != Plus stays unevaluated. (This is the best strategy since later definitions could make it decidable.)

Because the second argument of Condition (/;) does not evaluate to True, the pattern does not match.

Lesson to learn: For testing for structual inequality, use UnsameQ (=!=):

MatchQ[a, r_ /; Head[r] =!= Plus]

True

Of course, the same applies, mutatis mutandis, to Equal and SameQ.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, thanks! That helps a lot!
  – Jmeeks29ig
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

The problem is this:

If a is a symbol then its Head is Symbol, so Head[a] =!= Plus reduces to Symbol != Plus. Unequal (!=) is supposed to be a mathematical test for inequality. In this case, it just cannot decide whether Symbol != Plus should evaluate to True or to False, since both sides are Symbols. Here an example why this is undecidable with the current amount of information:

With[Symbol = 1, Plus = 1, Symbol != Plus]
With[Symbol = 1, Plus = 0, Symbol != Plus]

False

True

So the expression Symbol != Plus stays unevaluated. (This is the best strategy since later definitions could make it decidable.)

Because the second argument of Condition (/;) does not evaluate to True, the pattern does not match.

Lesson to learn: For testing for structual inequality, use UnsameQ (=!=):

MatchQ[a, r_ /; Head[r] =!= Plus]

True

Of course, the same applies, mutatis mutandis, to Equal and SameQ.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, thanks! That helps a lot!
  – Jmeeks29ig
  1 hour ago
  

  
  
  
  

add a comment | 

up vote
5
down vote



accepted

up vote
5
down vote



accepted

The problem is this:

If a is a symbol then its Head is Symbol, so Head[a] =!= Plus reduces to Symbol != Plus. Unequal (!=) is supposed to be a mathematical test for inequality. In this case, it just cannot decide whether Symbol != Plus should evaluate to True or to False, since both sides are Symbols. Here an example why this is undecidable with the current amount of information:

With[Symbol = 1, Plus = 1, Symbol != Plus]
With[Symbol = 1, Plus = 0, Symbol != Plus]

False

True

So the expression Symbol != Plus stays unevaluated. (This is the best strategy since later definitions could make it decidable.)

Because the second argument of Condition (/;) does not evaluate to True, the pattern does not match.

Lesson to learn: For testing for structual inequality, use UnsameQ (=!=):

MatchQ[a, r_ /; Head[r] =!= Plus]

True

Of course, the same applies, mutatis mutandis, to Equal and SameQ.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

The problem is this:

If a is a symbol then its Head is Symbol, so Head[a] =!= Plus reduces to Symbol != Plus. Unequal (!=) is supposed to be a mathematical test for inequality. In this case, it just cannot decide whether Symbol != Plus should evaluate to True or to False, since both sides are Symbols. Here an example why this is undecidable with the current amount of information:

With[Symbol = 1, Plus = 1, Symbol != Plus]
With[Symbol = 1, Plus = 0, Symbol != Plus]

False

True

So the expression Symbol != Plus stays unevaluated. (This is the best strategy since later definitions could make it decidable.)

Because the second argument of Condition (/;) does not evaluate to True, the pattern does not match.

Lesson to learn: For testing for structual inequality, use UnsameQ (=!=):

MatchQ[a, r_ /; Head[r] =!= Plus]

True

Of course, the same applies, mutatis mutandis, to Equal and SameQ.

share|improve this answer

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

share|improve this answer

share|improve this answer

edited 1 hour ago

edited 1 hour ago

edited 1 hour ago

answered 1 hour ago

Henrik Schumacher

38.4k251112

answered 1 hour ago

Henrik Schumacher

38.4k251112

answered 1 hour ago

Henrik Schumacher

38.4k251112

38.4k251112

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, thanks! That helps a lot!
  – Jmeeks29ig
  1 hour ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Awesome, thanks! That helps a lot!
  – Jmeeks29ig
  1 hour ago
  

  
  
  
  

Awesome, thanks! That helps a lot!
– Jmeeks29ig
1 hour ago

Awesome, thanks! That helps a lot!
– Jmeeks29ig
1 hour ago

add a comment | 

 

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