Hypothetical syllogism: is p→q, q→r ⊨ p→r valid?
Two conditionals sharing a middle formula chain into one: from p → q and q → r you get p → r. This is how short steps build a long argument.
Valid
p→q, q→r ⊨ p→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: q→rpremise
- 3False: p→rnegated conclusion
- 4True: pfrom line 3
- 5False: rfrom line 3
- 6False: pfrom line 1
Branch closed: line 6 contradicts line 4.
- 7True: qfrom line 1
- 8False: qfrom line 2
Branch closed: line 8 contradicts line 7.
- 9True: rfrom line 2
Branch closed: line 9 contradicts line 5.
closed branch