Using notI in Welder. Example contradiction proof - leon

Question

I have some problems while using the notI construct in Welder. Take the following as an example proof:
My example uses usual lemmas about the structure of rings and a derived lemma (zeroDivisorLemma) that says that for all 'a' in the ring a0 = 0 = 0a.
I proof that if two elements are not zero, their product is not zero. As follows.
val theorem: Theorem =
forallI("a"::F,"b"::F){ case (a,b) =>
implI(and(a !== z, b !== z,multInverseElement,multNeutralElement,multAssociative,
addOppositeElement,addNeutralElement,addAssociative,isDistributive)){ axioms =>
val Seq(aNotZero,bNotZero,multInverseElement,multNeutralElement,multAssociative,
addOppositeElement,addNeutralElement,addAssociative,isDistributive) = andE(axioms) : Seq[Theorem]
notI((a ^* b) === z) { case (hyp,_) =>
((a ^* b) === z) ==| andI(bNotZero,hyp) |
(((a ^* b) ^* inv(b)) === (z ^* inv(b))) ==| forallE(multAssociative)(a,b,inv(b)) |
((a ^* (b ^* inv(b))) === (z ^* inv(b))) ==| andI(forallE(multInverseElement)(b),bNotZero) |
((a ^* one) === (z ^* inv(b))) ==| forallE(multNeutralElement)(a) |
(a === (z ^* inv(b))) ==| forallE(implE(zeroDivisorLemma)(g => axioms))(inv(b)) |
(a === z) ==| aNotZero |
((a !== z) && (a === z)) ==| trivial |
(a !== a) ==| trivial |
BooleanLiteral(false)
}
}
}
The code compiles but Welder says:
SMT solver could not prove the property: false
from hypotheses:
(mult(a, b) == zero()) == false.
I'd say I'm not passing the right function to the construct. Could someone explain what should I write to succeed? Does it have to do with the type, i.e., (Theorem, Goal) => Attempt[Witness]? Do I need to provide a theorem and that proves the goal?
What else can I proof to get false? Should I use some sort of implication introduction?

Answer 1

What the error says is that it could not deduce a contradiction from what you shown. Indeed, you have not shown a contradiction in body of the notI. The derivation to prove that (mult(a, b) == zero()) == false is correct, but you still have to explicitly show the contradiction by taking the conjunction with hyp.
Would something like this work ?
val theorem: Theorem =
forallI("a"::F,"b"::F){ case (a,b) =>
implI(and(a !== z, b !== z,multInverseElement,multNeutralElement,multAssociative,
addOppositeElement,addNeutralElement,addAssociative,isDistributive)){ axioms =>
val Seq(aNotZero,bNotZero,multInverseElement,multNeutralElement,multAssociative,
addOppositeElement,addNeutralElement,addAssociative,isDistributive) = andE(axioms) : Seq[Theorem]
notI((a ^* b) === z) { case (hyp,_) =>
val deriv = ((a ^* b) === z) ==| andI(bNotZero,hyp) |
(((a ^* b) ^* inv(b)) === (z ^* inv(b))) ==| forallE(multAssociative)(a,b,inv(b)) |
((a ^* (b ^* inv(b))) === (z ^* inv(b))) ==| andI(forallE(multInverseElement)(b),bNotZero) |
((a ^* one) === (z ^* inv(b))) ==| forallE(multNeutralElement)(a) |
(a === (z ^* inv(b))) ==| forallE(implE(zeroDivisorLemma)(g => axioms))(inv(b)) |
(a === z) ==| aNotZero |
((a !== z) && (a === z)) ==| trivial |
(a !== a) ==| trivial |
BooleanLiteral(false)
andI(deriv, hyp)
}
}
}