Let's break down the problem step by step. ## Step 1: Understanding the Claim **Claim:** No sentence of SL (Sentential Logic) that contains **only binary connectives** is **TF-True** (tautologically true; true under all truth assignments). - **Binary connectives:** ∧ (and), ∨ (or), → (if...then), ↔ (if and only if) - **TF-True:** True under every possible assignment of truth values to propositional variables **Restated:** The claim says it is *impossible* to make a tautology in SL using only binary connectives. --- ## Step 2: Find a Counterexample Let's try to find a sentence using **only binary connectives** that **is** a tautology. ### Example: Consider the sentence: \[(P \lor \neg P)\] But here, **negation (¬)** is a unary connective. The question wants sentences with *only* binary connectives. So, we need to express negation using only binary connectives, which seems tricky. But perhaps the question means "all connectives are binary" (there is no use of unary connectives at all — i.e., negation is not allowed). #### Let's try another example using only binary connectives, no negation: - \[(P \rightarrow P)\] - This uses only the binary connective →. Let's build the truth table: | P | P → P | |---|-------| | T | T | | F | T | In both cases, the sentence is **true**. Therefore: **(P → P)** is a tautology using only a binary connective. --- ## Step 3: Demonstrate the Claim is False The claim is **false** because the sentence **(P → P)** contains only the binary connective →, and it **is TF-True** (a tautology). --- ## Step 4: Where an Inductive Proof Would Fail Suppose someone tries to **prove the original claim by induction**: - **Base case:** For atomic sentences (no connectives), obviously not TF-True, so OK. - **Inductive step:** Suppose all sentences of length ≤ n using only binary connectives are not tautologies. Consider a sentence of length n+1, built as (A ⋄ B) for some binary connective ⋄. They would try to argue that this can't be a tautology because neither A nor B can be tautologies by the inductive hypothesis. **But here's where it fails:** The sentence **(P → P)** is a sentence of SL using only one binary connective, and is tautological. This is a base case that is already a tautology, so the inductive hypothesis is **not true for all base cases**. Therefore, the inductive proof cannot proceed. The base case fails. --- ## Step 5: Final Answer ### **Counterexample:** \[ (P \rightarrow P) \] This sentence uses only the binary connective "→" and is TF-True (a tautology). ### **Explanation of Induction Failure:** An attempt at an inductive proof would fail at the **base case**, because sentences like \((P \rightarrow P)\) (using only a binary connective) are already tautologies. Therefore, the inductive hypothesis does not hold for all base cases, and the proof cannot go through. This demonstrates the claim is false.