15 / 18 · Account for every step
Cases, contradictions, and what they prove
Close every case and identify exactly which assumption a contradiction rejects.
Builds on Borrow an assumption, then return it
Go to practice ↓A question to keep in mind
A plan works in one of two cases. Is that a proof it always works?
From P ∨ Q, derive R by showing R under a temporary P assumption and also under a temporary Q assumption. Disjunction elimination (∨E) discharges both branches. You cannot simply choose the branch you prefer, and the cases must cover the possibilities asserted by the premise.
For negation introduction (¬I), assume P and derive a contradiction, written ⊥, such as Q and ¬Q together. Close that subproof and conclude ¬P. The contradiction rejects the added assumption relative to the other premises; it does not prove that every earlier statement is false.
In classical proof by contradiction, assuming ¬P and deriving ⊥ yields ¬¬P, then P by double-negation elimination. Other logical systems may not permit the last rule. We keep the classical assumption explicit.
Work through an example
- Premises: P ∨ Q, P → R, Q → R. P-branch: assume P, then R. Q-branch: assume Q, then R.
- Both covered cases reach R, so ∨E gives R outside both subproofs. The two cases need not be mutually exclusive.
- For P → Q and ¬Q, assume P. →E gives Q, which contradicts ¬Q. Closing the subproof gives ¬P by ¬I.
Your turn
0 / 3Complete this derivation of ¬P.
Read solution · does not award completion
- 3. [inside] P → Temporary assumption
- 4. [inside] Q → →E, 1 + 3
- 5. [inside] ⊥ → Contradiction, 2 + 4
- 6. [closed] ¬P → ¬I, discharge 3–5
The contradiction is Q with ¬Q. It closes the temporary P branch and establishes ¬P relative to the permanent premises.
You know P ∨ Q and P → R. Is that enough to infer R by cases?
Read solution · does not award completion
No, the Q case still needs a route to R.
A missing branch is a missing justification, not a formatting issue.
Construct the counterexample for the incomplete case analysis.
Read solution · does not award completion
P = F · Q = T · R = F
P = F, Q = T, R = F satisfies both premises and refutes R. This is exactly the uncovered branch.
Bring it back to your own work
Write a two-case plan. State why the cases cover the possibilities and what supports the desired outcome in each case.