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

score 878 · Answer 1 · 2024-12-26 13:30:00

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878

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

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score 396 · Answer 2 · 2024-12-26 13:13:14

Check for evars in a tactic that returns a value

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score 216 · Answer 3 · 2024-12-26 13:15:13

Proof of the application of a Substitution on a term

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Simplifying Subformulas in Coq

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score 624 · Answer 6 · 2024-12-26 13:26:20

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

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score 423 · Answer 7 · 2024-12-26 13:15:40

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

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score 453 · 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?

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score 154 · Answer 9 · 2024-12-26 13:26:21

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

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

score 740 · Answer 10 · 2024-12-26 13:24:31

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

740 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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