Q3: Yes, you should expect the axiom(s) and the goal(s) included as hints. Both of them can serve as useful. I more meant that you might see something like "$F" as a hint doesn't seem to add much to me, and that hints also lead you down a particular path...

recursion,isabelle,theorem-proving

The theoretical explanation First of all, the notion of a datatype of commands that allow non-deterministic choice from an arbitrary set of commands is deeply problematic. I will explain why. Suppose you had a datatype datatype Cmd = Skip | NonDeterministicChoice "Cmd set" like you wanted. Let A := (UNIV...

palindrome,coq,theorem-proving

This is a nice example where "direct" induction does not work well at all because you don't directly make the recursive call on the tail, but on part of the tail. In such cases, I usually advice to state your lemma with the length of the list, not on the...

Here is a proof that uses induction on n. Require Import NPeano. Theorem my_thm: forall n m, (n <? m) = false -> m <= n. induction n; destruct m; intros ; auto using (Le.le_n_S); discriminate. Qed. ...

linux,haskell,make,ghc,theorem-proving

Looks like paradox was written for a rather old version of GHC. You can fix all these "Could not find module" errors by using GHC version 7.8 or older and setting GHC = ghc -hide-package base -package haskell98 in the Makefile, though you will likely encounter more errors after that....

solver,isabelle,theorem-proving

There generally is no linear ordering on the strength of proving methods and your word "weakest" assume there is one. We can nevertheless say that "auto" generally have at least the power of "simp" or "rule" but as it is more powerful it can also do some useless work that...

proof,isabelle,theorem-proving,isar

It mostly depends on whether you are using the archaic (sorry for that ;)) apply-style or proper structured Isar for proving. I will give a small example to cover both styles. Assume you wanted to prove lemma "A & B" Where A and B just serve as placeholders for potentially...

constructor,agda,theorem-proving,disjoint-union

The data type constructors are disjoint. I'd say it's a theorem in Agda's type-system meta-theory. You can try to case the eq proof (C-c C-c), and Agda will find the contradiction: lemma : ∀ {a b} {A : Set a} {B : Set b} {x : A} {y : B}...

solver,isabelle,theorem-proving

I just asked Tobias Nipkow and this is what he told me: presburger is a decision procedure for Presburger arithmetic, i.e. linear arithmetic on natural numbers and integers, plus some pre-processing, which is why your statement with real could be proven as well (as it boils down to a problem...

sorting,z3,z3py,theorem-proving

This example is not well-formed SMT2, functions can not return multiple objects. The Z3 Guide for examples of how to use datatypes as well as quantifiers.

set,higher-order-functions,isabelle,theorem-proving,isar

HOL types cannot depend on values. So if you want to define a quotient type for an arbitrary non-empty set S and equivalence relation equiv using quotient_type, the arbitrary part must stay at the meta-level. Thus, S and equiv can either be axiomatized or defined such that you can convince...

Even though with default display settings, the subterm seems to appear in the goal, with Set Printing All enabled, it becomes clear that the the subterm does not match the goal because in the goal, stack has been unfolded to list nat. So fold stack is needed to turn list...

z3,topology,z3py,theorem-proving

It is possible that both a formula and the negation of the formula is consistent with respect to a background theory T. In particular, when T is not complete, then there are sentences that are neither consequences of T nor inconsistent with T. In your case the theory T is...

haskell,theorem-proving,dependent-type,higher-rank-types

I'm reading it as "given a term which requires l ~ r, return that term" It's "given a term whose type contains l, return that term with all ls being substituted by rs in the type" (or in the other direction r -> l). It's a very neat trick,...