Question #309512

1.1 Determine whether ( 𝑝∨𝑞)∧(𝑝→𝑟)∧( 𝑞→𝑠)→𝑟∨𝑠 is a Tautology or a contradiction

And give steps please


1
Expert's answer
2022-03-13T18:22:11-0400

Proof by contradiction

Suppose the formula is not a tautology.

Then there exist (p0,q0,r0,s0)∈(T,F)4(p_0,q_0,r_0,s_0 )∈(T,F)^4 such that

∣(p0⋁qo)⋀(p0→r0)∧(q0→s0)→r0∨s0∣=F|(p_0⋁q_o)⋀(p_0→r_0)∧(q_0→s_0)→r_0∨s_0 |=F .

The definition of implication implies that

∣(p0⋁qo)⋀(p0→r0)∧(q0→s0)∣=T|(p_0⋁q_o)⋀(p_0→r_0)∧(q_0→s_0)|=T and ∣r0∨s0∣=F|r_0∨s_0 |=F

The definition of conjunction and disjunction imply that

∣p0⋁qo∣=∣p0→r0∣=∣(q0→s0)∣=T|p_0⋁q_o|=|p_0→r_0|=|(q_0→s_0)|=T and ∣r0∣=∣s0∣=F|r_0|=|s_0 |=F

It follows from ∣p0∣→F=∣q0∣→F=T|p_0 |→F=|q_0 |→F=T that ∣p0∣=∣q0∣=F|p_0 |=|q_0 |=F

Consequently ∣p0∨q0∣=F∨F=F|p_0∨q_0 |=F∨F=F and we have a contradiction with ∣p0∨q0∣=T|p_0∨q_0 |=T

Therefore, our assumption is not true, and we conclude that the statement is a TAUTOLOGY


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