Affirming the consequent: is p→q, q ⊨ p valid?
A fallacy: p → q with q leaves p open, because q can hold for reasons that have nothing to do with p. The open branch below gives the countermodel.
Invalid
p→q, q ⊨ pA branch stays open, so the premises can hold while the conclusion fails.
Proof (semantic tableau)
- 1True: p→qpremise
- 2True: qpremise
- 3False: pnegated conclusion
- 4False: pfrom line 1
Branch open.p = ⊥, q = ⊤
- 5True: qfrom line 1
Branch open.p = ⊥, q = ⊤
open branch
Countermodels
Under these assignments every premise is true but the conclusion is false, so the conclusion does not follow.
p = ⊥, q = ⊤