Constructive dilemma: is p∨q, p→r, q→r ⊨ r valid?
Whichever of p or q holds, r follows — so r follows outright. A disjunction plus a conditional from each side gives the shared conclusion.
Valid
p∨q, p→r, q→r ⊨ rEvery branch of the tableau closes, so nothing makes the premises true and the conclusion false at once.
Proof (semantic tableau)
- 1True: p∨qpremise
- 2True: p→rpremise
- 3True: q→rpremise
- 4False: rnegated conclusion
- 5True: pfrom line 1
- 6False: pfrom line 2
Branch closed: line 6 contradicts line 5.
- 7True: rfrom line 2
Branch closed: line 7 contradicts line 4.
- 8True: qfrom line 1
- 9False: pfrom line 2
- 10False: qfrom line 3
Branch closed: line 10 contradicts line 8.
- 11True: rfrom line 3
Branch closed: line 11 contradicts line 4.
- 12True: rfrom line 2
Branch closed: line 12 contradicts line 4.
closed branch