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How do you selectively simplify arguments to each time a function is called, without evaluating the function itself?

score 857 · Answer 1 · 2024-12-26 13:30:00

2

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857

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How do you selectively simplify arguments to each time a function is called, without evaluating the function itself?

857 views Asked by scubed At 26 December 2024 at 13:30

score 380 · Answer 2 · 2024-12-26 13:13:14

Check for evars in a tactic that returns a value

380 views Asked by Rand00 At 26 December 2024 at 13:13

score 199 · Answer 3 · 2024-12-26 13:15:13

Proof of the application of a Substitution on a term

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score 1218 · Answer 4 · 2024-12-26 13:02:29

Coq contradiction in hypotheses

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score 521 · Answer 5 · 2024-12-31 00:49:20

Simplifying Subformulas in Coq

521 views Asked by Rand00 At 31 December 2024 at 00:49

score 603 · Answer 6 · 2024-12-26 13:26:20

is there any tactic in Coq that can transform a bool expression to a Prop one?

603 views Asked by Dan Johnson At 26 December 2024 at 13:26

score 402 · Answer 7 · 2024-12-26 13:15:40

Coq error: Unable to unify "true" with "is_true (0 < a - b - 3)"

402 views Asked by Dan Johnson At 26 December 2024 at 13:15

score 436 · Answer 8 · 2024-12-26 13:30:44

If two constructor expressions of an inductive type are equal in Coq, can I do rewriting based on their corresponding arguments?

436 views Asked by Dan Johnson At 26 December 2024 at 13:30

score 139 · Answer 9 · 2024-12-26 13:26:21

Can I use destruct here given the constraint I have for index range of a list?

139 views Asked by Dan Johnson At 26 December 2024 at 13:26

score 724 · Answer 10 · 2024-12-26 13:24:31

Take a conjunction of two hypotheses and create a new hypothesis in Coq

724 views Asked by Dan Johnson At 26 December 2024 at 13:24

TechQA.

List Question

How do you selectively simplify arguments to each time a function is called, without evaluating the function itself?

Check for evars in a tactic that returns a value

Proof of the application of a Substitution on a term

Coq contradiction in hypotheses

Simplifying Subformulas in Coq

is there any tactic in Coq that can transform a bool expression to a Prop one?

Coq error: Unable to unify "true" with "is_true (0 < a - b - 3)"

If two constructor expressions of an inductive type are equal in Coq, can I do rewriting based on their corresponding arguments?

Can I use destruct here given the constraint I have for index range of a list?

Take a conjunction of two hypotheses and create a new hypothesis in Coq

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