Negation introduction How do we prove a negation? P = P F -introduction {Assume} P (l-1) F { -intro on and (l-1)} (l) P -intro
Negation elimination -elimination How do we use a negation in a proof? P (l) P { -elim on and (l)} (m) F (k < m, l < m) P P F time for an example
F introduction F-introduction How do we prove F? P (l) P {F-intro on and (l)} (m) F P P F the same as -elim only intended bottom-up (k < m, l < m)
F elimination F-elimination How do we use F in a proof? it s very useful F P (m) F {F-elim on } P (k < m)
Double negation introduction -introduction How do we prove? P P (m) P { -intro on } P (k < m)
Double negation elimination -elimination How do we use in a proof? P P (m) P { -elim on } P (k < m)
Proof by contradiction Theorem Proof If x 2 is even, then x is even (x Z). Let x Z Assume x 2 is even. Assume that x is odd. Then x = 2y+1 for some y Z. Then x 2 = (2y+1) 2 = 4y 2 + 4y + 1 = 2(2y 2 + 2y) + 1 and 2y 2 + 2y Z. So, x 2 is odd a contradiction. So, x is even if y is not free in P and Q (sub)goal generating hypothesis pure hypothesis conclusion Thanks to Bas Luttik
Proof by contradiction proof by contradiction How do we prove P by a contradiction? {Assume} P P F P P (l-1) (l) (l+1) F { -intro on and (l-1)} P { -elim on (l)} P (k < m) -intro -elim time for an example
Disjunction introduction -introduction How do we prove a disjunction? {Assume} P (l-1) Q { -intro on and (l-1)} (l) P Q P Q P Q Q P P Q -intro
Disjunction introduction -introduction How do we prove a disjunction? {Assume} Q (l-1) P { -intro on and (l-1)} (l) P Q P Q P Q Q P P Q -intro
Disjunction elimination -elimination How do we use a disjunction in a proof? P Q P Q P Q Q P (m) P Q { -elim on } P Q (k < m)
Disjunction elimination -elimination How do we use a disjunction in a proof? P Q P Q P Q Q P (m) P Q { -elim on } Q P (k < m)
Proof by case distinction How do we prove R by a case distinction? proof by case distinction P Q (P Q) (P R) (Q R) R (l) P R (m) Q R (n) {case-dist on, (l), (m)} R (k < n, l< n, m<n)
Bi-implication introduction How do we prove a bi-implication? (P Q) (Q P) P Q -introduction (l) P Q Q P -intro (m) { -intro on and (l)} P Q (k < m, l < m)
Bi-implication elimination How do we use a bi-implication in a proof? -elimination P Q (P Q) (Q P) P Q P Q { -elim on } { -elim on } -elim (m) P Q (m) Q P (k < m) (k < m)
Derivations / Reasoning with quantifiers
Proving a universal quantification To prove x[x Z x 2 : x 2-2x 0] Proof Let x Z be arbitrary and assume that x 2. Then, for this particular x, it holds that x 2-2x = x(x-2) 0 (Why?) Conclusion: x[x Z x 2 : x 2-2x 0]. if y is not free in P and Q
introduction -introduction How do we prove a universal quantification? {Assume} var x; P(x) (l-1) Q(x) { -intro on and (l-1)} (l) x[p(x) : Q(x)] similar to -intro with generating hypothesis flag shows the idity of a hypothesis
Using a universal quantification We know x[x Z x 2 : x 2-2x 0] Whenever we encounter an a Z such that a 2, we can conclude that a 2-2a 0. For example, (52387 2-2 52387) 0 since 52387 Z and 52387 2.
elimination How do we use a universal quantification in a proof? -elimination similar to implication but we need a witness time for an example (l) (m) x[p(x) : Q(x)] P(a) { -elim on and (l)} Q(a) (k < m, l < m) a is an object (variable, number,..) which is known in line (l) the same a from line (l)
introduction -introduction How do we prove an existential quantification? {Assume} x[p(x) : Q(x)] (l-1) F { -intro on and (l-1)} (l) x [P(x) : Q(x)] x[p(x): Q(x)] x [P(x) : Q(x)] and -intro
elimination How do we use an existential quantification in a proof? -elimination x [P(x) : Q(x)] x[p(x): Q(x)] (l) (m) x [P(x) : Q(x)] x[p(x): Q(x)] { -elim on and (l)} F (k < m, l < m) and - elimination time for an example
Proofs with -introduction and - elimination are unnecessarily long and cumbersome There are alternatives
Alternative elimination How do we use an existential quantification in a proof? we pick a witness (m) x [P(x) : Q(x)] { *-elim on } Pick x with P(x) and Q(x) (k < m) x must be new time for an example