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Order Now1. [1 marks] Let X ⊆ Φ and α ∈ Φ. Prove that X α iff X ` α.
2. Let X be a set of formulas. X is said to be a finitely satisfiable set (FSS) if every Y ⊆f in X is
satisfiable.
Equivalently, X is an FSS if there is no finite subset {α1, α2, . . . , αn} of X such that ¬(α1∧α2∧. . .∧αn)
is valid.
(Note that if X is an FSS we are not promised a single valuation v which satisfies every finite subset
of X. Each finite subset could be satisfied by a different valuation.)
Show that:
(a) [0.5 marks] Every FSS can be extended to a maximal FSS.
(b) [0.5 marks] If X is a maximal FSS then for every formula α, α ∈ X iff ¬α /∈ X.
(c) [0.5 marks] If X is a maximal FSS then for all formulas α, β, (α∨β) ∈ X iff (α ∈ X or β ∈ X).
(d) [1 marks] Every maximal FSS X generates a valuation vX such that for every formula α,
vX α iff α ∈ X.
From these facts, conclude that:
(e) [0.5 marks] Any FSS X is simultaneously satisfiable (that is, for any FSS X, there exists vX
such that vX X).
(f) [1 marks] For all X and all α, X α iff there exists Y ⊆f in X such that Y α.
3. [2 marks] Recall the coding exercise from Assignment 1. Assume that we only have unsatisfiable
formulas as inputs, in our formula files. Suppose we modify the proof file of our sample input (from
Assignment 1) as follows:
3f 4f
4f 5f
1p 2p
6f 3p
This modified proof, as you can see, only contains the clause-ids in each line. We can check the
correctness of this proof, under the assumption that once the clauses (line numbers) have been identified,
the corresponding resolvent is correct and unique. For uniqueness, assume that if there are more than
one pair of complementary literal, resolution is done by picking the pair corresponding to the smaller
atom, i.e. the pair (1, -1) should be used before (2, -2), and so on.
Correctness of such a proof essentially depends on three things:
• the clauses indicated by the clause ids in each line of proof file indeed contains (at least) a pair
of complementary literals,
• the clauses of the last line of the proof file must give an empty clause as resolvent, and
• the clause ids in each line in the proof file only refers to clauses that are in the formula file, or
already discovered clauses in the proof file.
Here’s what you have to do. Given a formula and a modified proof file (like the one shown above)
but with holes (denoted by “??”), you need to output a complete proof file (like the sample proof file
shown in Assignment 1). Write a Python code for this task. You are guaranteed that there will be at
most one hole (in place of only one of the clause ids) in every line of the input proof. If the proof with
holes cannot be completed anyhow, return the proof file with every “??” replaced with “np”.
Here’s a sample formula:
c CNF formula (p1 ∨ !p2) ∧ (p2 ∨ p3) ∧ (!p1 ∨ !p2 ∨ p3) ∧ (!p3)
p cnf 3 4
1 -2 0
2 3 0
-1 -2 3 0
-3 0
And, here is a sample proof with holes:
3f ??
?? 5f
1p ??
?? 3p
And here is the sample output:
3f 4f 1 3 0
4f 5f -1 3 0
1p 2p 3 0
6f 3p 0
(This is not a part of the assignment, but note that if you are able to output a complete proof, you can
verify the correctness of your output using the code that you had submitted for Assignment 1, assuming
that the code was correct!)
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