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Joined 2026-09-21 Message
Incoming links: Disjunction elimination rule
Assumption discharge 2026-10-25
Assumption discharge closes a designated temporary assumption together with its dependent subproof. Implication introduction rule turns a logical derivation of under into outside that temporary scope. Negation introduction rule turns a logical derivation of logical falsity under into . Disjunction elimination rule discharges its separate case assumptions while retaining any surrounding premises.
Logical falsity elimination rule 2026-10-25
From logical falsity infer any proposition. This rule permits a contradictory case inside a disjunction elimination rule proof to reach the common required conclusion.
Apply the complete logical derivation of the law of excluded middle in (a) to , obtaining . Then use the disjunction elimination rule, constructing each logical implication by an explicit hypothetical subproof:
1 Q ∨ ¬Q result of (a), φ := Q
2 | Q first case
3 | | P assume
4 | | Q available from 2
5 | P → Q →-intro, discharge 3–4
6 | (P → Q) ∨ (P → ¬Q) ∨-intro-L, 5
7 | ¬Q second case
8 | | P assume
9 | | ¬Q available from 7
10 | P → ¬Q →-intro, discharge 8–9
11 | (P → Q) ∨ (P → ¬Q) ∨-intro-R, 10
12 (P → Q) ∨ (P → ¬Q) ∨-elim, 1, cases 2–6 and 7–11Both case assumptions and both hypothetical assumptions have been discharged. Thus the conclusion is a theorem, not a logical implication that depends on assuming .
The logical implication associates to the right. Assume its two antecedents, then use nested applications of the disjunction elimination rule.
1 | P ∨ Q assume
2 | | P ∨ ¬Q assume
3 | | | P first case of 1
4 | | | P available from 3
5 | | | Q second case of 1
6 | | | | P first case of 2
7 | | | | P available from 6
8 | | | | ¬Q second case of 2
9 | | | | ⊥ ¬-elim, 5 and 8
10 | | | | P ⊥-elim, 9
11 | | | P ∨-elim, 2, cases 6–7 and 8–10
12 | | P ∨-elim, 1, cases 3–4 and 5–11
13 | (P ∨ ¬Q) → P →-intro, discharge 2–12
14 (P ∨ Q) → ((P ∨ ¬Q) → P) →-intro, discharge 1–13This proof uses logical falsity elimination rule, but does not need law of excluded middle or double-negation elimination.
Expand three cases as two applications of disjunction elimination rule. For the association , the first outer branch assumes and applies the given and subproofs to obtain . The other outer branch is the given subproof. In schematic natural deduction form:
1 (φ ∨ ψ) ∨ ρ given
2 | φ ∨ ψ first outer case
3 | | φ first inner case
4 | | κ given φ-subproof
5 | | ψ second inner case
6 | | κ given ψ-subproof
7 | κ ∨-elim, 2, cases 3–4 and 5–6
8 | ρ second outer case
9 | κ given ρ-subproof
10 κ ∨-elim, 1, cases 2–7 and 8–9All three case assumptions are discharged and common surrounding assumptions remain. If the original three-way logical disjunction is parsed as , use the subproof as the first outer branch and eliminate inside the second. Thus either binary association has the asserted three-case disjunction elimination, without presupposing an unproved associativity.
- The implication introduction rule assumes locally, derives under that assumption, then concludes while discharging the temporary assumption.
- The disjunction elimination rule uses and two local cases: one derives a common conclusion from , and the other derives that same from . It concludes and discharges both case assumptions.
- The negation introduction rule assumes locally and derives logical falsity. Discharging that assumption gives .
The boxes delimit a subproof and identify the lifetime of its temporary assumptions. A result depending on a boxed assumption is available inside that box and its nested boxes; it cannot simply be exported outside. The enclosing rule specifies what may be exported and which assumptions are discharged. Other premises on which the derivation depends remain undischarged. This assumption discharge discipline distinguishes a conditional argument from an unconditional proof.
Assume the conjunction and obtain a contradiction in each of the two cases supplied by the premise:
| Line | Formula | Justification |
|---|---|---|
| 1 | Premise | |
| 2 | Assume locally | |
| 3 | Conjunction elimination rule, left, 2 | |
| 4 | Conjunction elimination rule, right, 2 | |
| 5 | Assume first case | |
| 6 | Negation elimination rule, 3 and 5 | |
| 7 | Assume second case | |
| 8 | Negation elimination rule, 4 and 7 | |
| 9 | Disjunction elimination rule, 1, 5–6 and 7–8 | |
| 10 | Negation introduction rule, discharging 2–9 |
Let abbreviate , and group the three equality alternatives on the left. The proof uses the binary disjunction elimination rule twice, so no unproved three-way derived rule is needed.
| Line | Formula | Justification |
|---|---|---|
| 1 | Premise | |
| 2 | Assume for the first implication | |
| 3 | Assume for the second implication | |
| 4 | Assume with fresh witness name | |
| 5 | Universal quantifier elimination rule, 1 | |
| 6 | Assume the first outer case | |
| 7 | Assume the first inner case | |
| 8 | Equality elimination rule, 7 and 4 | |
| 9 | Negation elimination rule, 2 and 8 | |
| 10 | Logical falsity elimination rule, 9 | |
| 11 | Assume the second inner case | |
| 12 | Equality elimination rule, 11 and 4 | |
| 13 | Disjunction introduction rule, left, 12 | |
| 14 | Disjunction elimination rule, 6, 7–10 and 11–13 | |
| 15 | Assume the second outer case | |
| 16 | Equality elimination rule, 15 and 4 | |
| 17 | Disjunction introduction rule, right, 16 | |
| 18 | Disjunction elimination rule, 5, 6–14 and 15–17 | |
| 19 | Existential quantifier elimination rule, 3 and 4–18 | |
| 20 | Implication introduction rule, discharging 3–19 | |
| 21 | Implication introduction rule, discharging 2–20 |
The witness name is absent from and from every assumption remaining after line 19, so the existential elimination is legal. Each equality is used in the forward direction supplied by the printed equality rule. The three named constants need not denote distinct objects: the case proof requires only that they exhaust the domain. This is existential quantification over a finitely named domain followed by exclusion of the first case.
Use the disjunction elimination rule on the outer disjunction. In its left branch, assume and apply the disjunction elimination rule again: from a case assumption , the disjunction introduction rule gives ; from a case assumption , the premise and the modus ponens rule give , then the disjunction introduction rule gives . Discharging those two inner case assumptions yields under the left outer branch.
In the right outer branch, assume . A case assumption gives by the disjunction introduction rule. A case assumption , together with the premise , gives by the negation elimination rule, and then by logical falsity elimination rule. Discharge these inner case assumptions with the disjunction elimination rule.
Both outer branches now have the same conclusion. Discharge the two outer case assumptions with the disjunction elimination rule to obtain under exactly the stated premises.
Keep the four given premises in context and assume under label . The modus ponens rule applied to the first two premises gives and . Apply the disjunction elimination rule to as follows.
Under a case assumption , the premise gives by the modus ponens rule. Together with , the negation elimination rule gives , and the supplied logical falsity elimination rule gives . Under the other case assumption , the required conclusion is already available. Discharge both case assumptions with the disjunction elimination rule to conclude .
Finally discharge with the implication introduction rule, giving . The premise is not needed: a derivation is permitted to leave a premise unused.
Assume under label , under label , and under label . A temporary case assumption gives by the modus ponens rule using . A temporary case assumption gives by the modus ponens rule using . The disjunction elimination rule discharges these case assumptions and concludes under .
Apply the implication introduction rule three times, discharging , then , then . The resulting theorem isThe displayed parentheses make explicit the right association of the source formula.
Write . We derive the two disjunction introduction rules and the disjunction elimination rule without using any disjunction rule inside the derivations.
For left introduction, start with and temporarily assume . The conjunction elimination rule gives , and the negation elimination rule gives . Discharge the temporary conjunction by the negation introduction rule, concluding . For right introduction, start with and use the right conjunction elimination rule to obtain instead. Thus either input derives the required encoded disjunction.
For elimination, suppose is available, together with a derivation of from an additional assumption and a derivation of from an additional assumption . Keep their common external context fixed. Temporarily assume under label . Temporarily assume under label and insert the supplied first branch derivation to obtain . Negation elimination rule gives , so negation introduction rule discharges and gives . Repeat with a temporary assumption under label and the supplied second branch derivation, discharging to obtain .
Conjunction introduction rule now gives . Its negation gives by negation elimination rule. The supplied reductio ad absurdum rule discharges and yields . This discharges the two branch assumptions as required and proves the encoded elimination rule. Its last step uses classical propositional logic; the same encoded elimination principle is not generally available in Intuitionistic propositional logic.
Let be fresh, not free in any of the three universal premises. Universal quantifier elimination rule gives , and . Apply disjunction elimination rule to the last formula.
In the branch, apply disjunction elimination rule to . Its case contradicts by negation elimination rule, and logical falsity elimination rule then gives . Its case already gives that conclusion. Discharge those two inner case assumptions to obtain under the outer assumption. In the outer branch, the modus ponens rule applied to gives directly.
Discharge the two outer case assumptions by disjunction elimination rule. The conclusion now depends only on the universal premises, in which is not free. Universal quantifier introduction rule therefore gives . All case assumptions have been discharged, and the argument requires no classical reductio ad absurdum.
Choose arbitrary fresh variables and assume under label . The conjunction elimination rule gives and . Apply the disjunction elimination rule to the equality disjunction.
In the case , the supplied symmetry rule gives , and the equality elimination rule applied to gives . Disjunction introduction rule gives . In the case , equality elimination rule applied directly to gives ; the other disjunction introduction rule gives the same conclusion . Discharge the two equality case assumptions by disjunction elimination rule.
Discharge with the implication introduction rule. There are now no undischarged assumptions. Use the universal quantifier introduction rule successively for , then , then , yieldingFreshness holds at every generalization because the sole temporary assumption has already been discharged; the arbitrary variables are not tied to any premise.
Three-case disjunction elimination 2026-10-25
If each of , and separately yields the same conclusion , then yields , with the three case assumptions discharged. For , first eliminate the outer logical disjunction, using a nested applications of the disjunction elimination rule on the branch. For , nest the second elimination in the other branch. Common surrounding assumptions remain open.