As we have to prove no pit in room [1,2] & [2,1], so we need to follow only those rules containing pit for only (1,2) and (2,1) and breeze for only (1,1), i.e. rule 2 and rule 4 only. R2 : B 1,1 ↔ (P 1,2 ˅ P 2,1 ), means if there is breeze in (1,1) then pit in either (1,2) or (2,1) and if pit in either 1,2 or 2,1 then breeze in (1,1) can be written in formal language as, B 1,1 → (P 1,2 ˅ P 2,1 ) & (P 1,2 ˅ P 2,1 ) → B 1,1 By And-Elimination rule to above logical representation by eliminating the term after &,we get B 1,1 → (P 1,2 ˅ P 2,1 ) R4 : ¬ B 1,1 no breeze in [1,1] Using above rule, B 1,1 → (P 1,2 ˅ P 2,1 ) can be represented as ¬ B 1,1 → ¬(P 1,2 ˅ P 2,1 ) By MP rule to above logical representation, can be represented as ¬(P 1,2 ˅ P 2,1 ) By applying De Morgan`s rule to the above logical representation, ¬(P 1,2 ˅ P 2,1 ) → ¬P 1,2 ˄ ¬ P 2,1 ¬P 1,2 ˄ ¬ P 2,1 ; Hence it is proved, no pit in [1,2] & [2,1],