Question

Consider this principle, the Beaver analogue of the validity "if φ, then φ": σ[φ implies φ]σ

L0:{atom, not, and, implies} L1: {atom, not, and, implies, ∂} L2:{atom,not,and,implies,□,♢}

Prove IMPLIES ID holds in L0. Prove that it does not hold in L2 by generating a counter example. Determine if IMPLIES ID holds in L1. If it does hold, prove it. If it does not, provide a counterexample.

Answer 1

The IMPLIES ID principle is analyzed in three logical systems: L0, L1, and L2. The validity of IMPLIES ID is proven in L0, shown to not hold in L2 through a counterexample, and the status in L1 is not determined yet.

Step-by-step explanation:

The IMPLIES ID principle states that if φ implies φ, then it holds true in a given logical system. Let's examine whether IMPLIES ID holds in the given logical systems:

1. L0: We have the logical connectives {atom, not, and, implies}. In this system, we can prove IMPLIES ID by considering the possible truth values for φ and checking if φ implies φ holds true. For example, if φ is true, then φ implies φ is true, and if φ is false, then φ implies φ is also true.
2. L1: In this system, the logical connectives are expanded to {atom, not, and, implies, ∂}. To determine whether IMPLIES ID holds, we need to check if φ implies φ holds true for all possible truth values of φ.
3. L2: This system includes additional logical connectives {atom, not, and, implies, □, ♢}. To show that IMPLIES ID does not hold, we can generate a counterexample where φ implies φ is false.