G4ip is a sequent calculus for propositional logic (by Dyckhoff) that is contraction-free, thus (if I understand correctly) greatly simplifying the writing of automated theorem provers by avoiding loops.
The original sequent calculus (LJ by Gentzen) is made for first-order logic, but is not contraction free.
The GI calculus (presented here p. 9) seems to have a rule similar to contraction (rule $\forall\!\to$), but the paper does not seem to indicate the necessity of such rule.
Is it possible to have a contraction-free sequent calculus for FOL ? If so, how ?
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Worrying about contraction in first-order logic is like lamenting the fact that your food wasn't served in a spillproof container while your airplane stalls and crashes into the ground. Technically, spillproof containers would strictly improve the situation, but in practice so many other things are going wrong that it wouldn't make a difference.
Contraction-free calculi let us perform exhaustive proof search in propositional logic, in a particularly simple and controlled way. We will never have an exhaustive proof search algorithm for first-order logic, since first-order logic is undecidable (even when you restrict to unary predicates in the intuitionistic case). A contraction-free first-order calculus would not have much use.
Putting the uselessness of contraction-free first-order calculi aside, one still has a hard time answering the technical question you posed ("can we have in principle a sequent calculus for FOL that is contraction-free") without having a formal definition of sequent calculus and contraction-free sequent calculus.
E.g., I could propose the calculus System X which has one rule $\overline{\Gamma \vdash P}$ with no premises for each FOL-provable sequent $\Gamma \vdash P$. This "calculus" is sound and complete for FOL, and there's nothing that looks like a contraction rule. But people would argue that System X is not really a sequent calculus, and they'd have a good point.
I could then propose a variant of LJ which folds the structural rules into the logical rules, so that e.g. $\wedge L_1$ takes the form
$$\frac{\Gamma,A,A\wedge B \vdash \Delta}{\Gamma, A \wedge B \vdash \Delta} $$
instead of the usual
$$\frac{\Gamma,A \vdash \Delta}{\Gamma, A \wedge B \vdash \Delta} $$
and insist that this calculus is contraction-free, since you technically never need to use the contraction rule. Of course, this would be a frivolous claim on my part: this is not at all what people have in mind when they talk about contraction-free calculi! But the fact remains that most things in proof theory, including the term "contraction-free calculus", lack all-encompassing rigorous definitions. If you intend to ask a question about proof calculi, you have to fill in these details yourself, and the answer will depend on how you fill those details in!
Here's an example of how you could fill in the details and arrive at a question that can actually be answered.
Sample Question: Can we find
so that
every provable sequent $\Gamma \vdash \Delta$ of LJ has a proof using only rules from the set $F$,
the premises of each rule in $F$ have smaller $\nu$-complexity than the conclusion?
Note that, for propositional sequents, the set of rules of the contraction-free calculus LJT of Roy Dyckhoff constitute such a set $F$ when equipped with the complexity measure defined by the equations $\nu(A \wedge B) = 2 + \nu(A) + \nu(B)$, $\nu(A \vee B) = \nu(A \rightarrow B) = 1 + \nu(A) + \nu(B)$, $\nu(\bot) = 1$, $\nu(\Gamma \vdash \Delta) = \nu(\bigwedge \Gamma \rightarrow \bigvee \Delta)$ that takes the value $0$ on atomic formulae.
Sample Answer: Not in general. The existence of such a measure would allow you to put a priori bounds on the lengths of proofs of a given sequent, and would therefore imply that FOL is decidable. But I can't rule out the existence of such calculi for certain well-behaved subsystems of FOL (e.g. classical but not intuitionistic monadic FOL).