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Tableau for (T∨S)→¬Q

Short answer

The tableau for (T∨S)→¬Q branches: left produces ¬(T∨S) giving ¬T, ¬S (open); right gives ¬Q (open). No branch closes, so the formula is satisfiable but not valid. Models: {T=false, S=false} or {Q=false}.

Draw the tableau for the formula (T∨S) → ¬Q where ¬ denotes the negation of variable, ∨ denotes the disjunction of variables and → is the symbol for implication.

Formula: (T ∨ S) → ¬Q.

Rule used. A → B is a β-formula: it branches into ¬A | B. Then ¬(T ∨ S) is an α-formula: it gives ¬T and ¬S on the same branch.

Tableau

          (T ∨ S) → ¬Q
            /      \
     ¬(T ∨ S)       ¬Q
         |           o
        ¬T
         |
        ¬S
         o

Reading the tableau

BranchLiteralsClosed?
Left¬T, ¬SNo, open (o)
Right¬QNo, open (o)

Neither branch contains a complementary pair, so the tableau is open and the formula is satisfiable.

  • Left branch gives the model T = false, S = false (Q can be anything).
  • Right branch gives the model Q = false (T, S can be anything).

Since the tableau is open, the formula is not unsatisfiable. (It is not valid either, for example T = true, Q = true makes it false.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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