Determine whether ( 𝑝∨𝑞)∧(𝑝→𝑟)∧( 𝑞→𝑠)→𝑟∨𝑠 is a Tautology or a contradiction using a truth table
Give full truth table
Solution
The truth table for (p∨q)∧(p→r)∧(q→s)→r∨s(p∨q) ∧(p→r) ∧(q→s) → r∨s(p∨q)∧(p→r)∧(q→s)→r∨s is shown below
From the last column, we can see that, all TRUE
Therefore,
(p∨q)∧(p→r)∧(q→s)→r∨s(p∨q) ∧(p→r) ∧(q→s) → r∨s(p∨q)∧(p→r)∧(q→s)→r∨s is a Tautology
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