Suppose a calculus had three rules:
Γ⇒φ,∆ Γ⇒φ◦ψ,∆ R◦1 Γ⇒ψ,∆ Γ⇒φ◦ψ,∆ R◦2 Γ⇒φ, ψ,∆ Γ⇒φ◦ψ,∆ R◦3
R◦3is invertible1but it is not shown invertible by our lemmata. Suppose a derivation ended
with Γ⇒∆⊕A◦B. It could have come from an instance of any of the three rules above. In the former two cases, we simply weaken with the appropriate formula in the premisses (B and A respectively), and in the latter case the result is immediate. R◦1 and R◦2 are
neither superfluous or redundant, however it would appear they are not needed.
The following definition can be used for any type of calculus, but we specialise it to decomposable uniprincipal calculi composed only of normal rules:
Definition 22 (Combinable rules) A pair of rules iscombinable iff they have the same conclusion AND only differ in one premiss. In other words, combinable right rules are of the form (left rules are an obvious analogy):
Γ,Γ1⇒∆,∆1 · · · Γ,Γi⇒∆,∆i · · · Γ,Γn⇒∆,∆n Γ⇒∆, ?s(Φ)~
Γ,Γ1⇒∆,∆1 · · · Γ,Γ0i⇒∆,∆0i · · · Γ,Γn⇒∆,∆n Γ⇒∆, ?s(Φ)~
A combined rule is created by replacing a pair of combinable rules with a rule which has the same conclusion AND:
• Where the combinable rules have identical premisses, the new rule has this premiss, AND
• Where the combinable rules have different premisses, the new rule has a premiss whose active part is the two active parts added together.
In other words, the combined rule from the pair of combinable rules above is: Γ,Γ1⇒∆,∆1 · · · Γ,Γi,Γ0i⇒∆,∆i,∆0i · · · Γ,Γn⇒∆,∆n
Γ⇒∆, ?s(Φ)~
a
We can generalise the definition to allow IW rules. If at least one of the rules to be combined is an IW rule, the combined rule is an IW rule.
Obviously, this technique is only valid where we can weaken on both sides of the sequent arrow. Lemma 7 (p. 81), and its associated results in different chapters, provides the sufficient conditions for a calculus to height preserving admit Weakening. The following result is similar to those of section5.2:
Lemma 17 (Removing Combinable Rules) LetRbe a multisuccedent calculus defined by a set of uniprincipal decomposable rules. LetR0 be a calculus based on Rwhere all pairs
of combinable rules have been replaced a combined rule. If a sequent was derivable in R, then it is derivable at the same height inR0.
Proof. Clearly we need only consider the cases where the final step in the derivation inR
was an instance of a rule in a combinable pair. We show the case where the rule is a right rule, the left case is symmetric. Suppose wlog that the premiss which ensured the rule was combinable was the first one. Then, the derivation ends with:
Γ,Γ1⇒∆,∆1 · · · Γ,Γn ⇒∆,∆n Γ⇒∆, ?s(B)~
Then, all of the premisses are derivable inR. Suppose the other rule in the combinable pair has an active part with first premiss Γ01 ⇒∆01. Weakening is depth-preserving admissible by lemma 7, hence we weaken the first premiss with Γ01 on the left and ∆01 on the right. Then:
Γ,Γ1⇒∆,∆1
Γ,Γ1,Γ01⇒∆,∆1,∆01 · · · Γ,Γn ⇒∆,∆n Γ⇒∆, ?s(B)~
is a derivation inR0. a
Again, the above proof considers normal rules only. If IW rules are considered, the proof is similar.
We then have a lemma similar to lemmata15(p. 93) and16(p. 94):
Lemma 18 (Removing Combinable rules) Let R be a multisuccedent calculus defined by a set of decomposable uniprincipal rules, and let R0 be obtained fromR by replacing all
combinable pairs with combined rules. If R ∈ R was shown invertible by a lemma from section6.2, andR is not part of a combinable pair, thenRin invertible in R0.
Proof. Similar to lemma15. a
When the calculusRadmitsContraction then removing combinable rules is conservative (in other words, no further sequents become provable). It is not known whether removing combinable rules from a calculus which does not admitContraction is conservative.
6.6.1
Example
We have now three results which will aid in proof search. Consider the following calculus with five rules for one ternary connective◦:
Γ⇒∆, φ1 Γ⇒∆,◦(φ1, φ2, φ3) R◦1 Γ⇒∆, φ2 Γ⇒∆,◦(φ1, φ2, φ3) R◦2 Γ⇒∆, φ1, φ2 Γ⇒∆, φ3 Γ⇒∆,◦(φ1, φ2, φ3) R◦3 Γ⇒∆, φ1 Γ, φ1⇒∆ Γ⇒∆, φ3 Γ,◦(φ1, φ2, φ3)⇒∆ L◦1 Γ, φ1, φ2⇒∆ Γ,◦(φ1, φ2, φ3)⇒∆ L◦2
Which rules are invertible? Only L◦2, and not by the methods of section 6.2. The left
premiss ofR◦3 is derivable at a height not greater than that of the conclusionR◦3.
Firstly,R◦1andR◦2form a combinable pair. Replace them with the combined ruleR◦0:
Γ⇒∆, φ1, φ2
Γ⇒∆,◦(φ1, φ2, φ3)
NowR◦3is redundant, so we can remove it. This leaves the ruleR◦0, which is invertible by
lemma8.
The calculus consisting of R◦0,L◦
1andL◦2 admits multi-cut, orCut for structures. In
other words, the following rule is admissible:
Γ⇒∆,Θ Γ,Θ⇒∆ Γ⇒∆ Cutm
where Θ is some multiset of formulae. From this, one can deduce that Cut (for single formulae) is admissible (see, for instance, [Shoesmith and Smiley, 2008]). Then, we can remove the superfluous ruleL◦1to leave a calculus with justR◦0 andL◦2as the only rules.
Using lemmata8 (p. 82) and9(p. 85) the calculus is invertible.