Propositions have truth values in an interpretation
Objective: Why is a false premise not a counterexample to p→q?
A proposition is a declarative statement that is true or false once its terms and context are fixed.
Objects and certificate: A proposition has one truth value in a fixed interpretation; ¬p, p∧q, p∨q, and p→q form new propositions by defined truth rules. The result to establish is: The implication p→q is false exactly when p is true and q is false. Decide whether its evidence is a truth table, witness, bijection, partition, induction chain, invariant, recurrence, graph trace, or normalized probability.
The worked question is: Evaluate (p→q)∧p when p=true and q=false. Begin with “p→q is false in the one violating row.” and complete the reconstruction with “false∧true is false.” Then compare the certificate with this boundary: “Open the door” is a command, not a proposition with a truth value.
The implication p→q is false exactly when p is true and q is false.
An implication promises q for every case in which p holds.
The row p=true,q=false violates that promise.
If p is false there is no active p-case, and if q is true the promised result holds.
Evaluate (p→q)∧p when p=true and q=false.
- p→q is false in the one violating row.
- p itself is true.
- false∧true is false.
Result: The compound proposition is false.