B.5.1 Propositional Consequences
holds independently of the interpretation of p and
consequently independently of the truth value of
(forall x: exists y: p(x, y)).
This is because this conclusion is an instance of the pattern
~~forall x: exists y: p(x,
forall x: exists y: p(x,
which is true independently of the truth value of G. We call such a
conclusion a propositional consequence (aussagenlogische Folge).
Proposition 153 (Propositional Consequences)
The following conclusions are propositional consequences for every formula
A and B:
- And Introduction and Or Elimination
- De Morgan
- Modus Ponens
There are (infinitely) many other propositional consequences.
A general strategy to show that
is a propositional consequence is to show that
A => B
is a propositional tautology.
Proposition 154 (Tautology)
A propositional formula with variables is a propositional
tautology (aussagenlogische Tautologie) if it is true for every assignment of
truth values to the variables.
We can show that a propositional formula with variables is a tautology by
constructing a truth table or by applying the indirect method.
We show that
((A \/ B) /\ (A => C) /\ (B => C)) => C.
is a tautology by assuming that its truth value is false and then
deriving a contradiction:
((A \/ B) /\ (A => C) /\ (B => C)) => C
Because the implication is false, C is false and the
conjuncts are true. Thus
A and B must be false (since C is false, the
implications cannot be true otherwise). Therefore A \/ B is false, which contradicts above derivation.
Author: Wolfgang Schreiner
Last Modification: October 4, 1999