(3 If and if, the above contradiction of cases (1) and (2), is also obtained by replacing.
On the contrary, we suppose that σ ( f ) = k ≥ 2. Thus, we will deduce a contradiction for cases A = ( − 1 ) n and A ≠ ( − 1 ) n, respectively.
We may get a contradiction for case 2 similar to case 1.
A P = { E ∗ }, A P = { E 0 }, A P is a closed orbit, or A P is a singular closed trajectory, where A P is negative limit set of L P. Case (i) and Lemma 3.1 lead to a contradiction, and case (ii) and Lemma 3.2 lead to a contradiction.
By analogous methods, we can obtain a contradiction for case.
Therefore, we get the contradiction in case that at least one of conditions (7), (8) holds which completes the proof.
This condition is required in order to avoid "Hegelian" contradictions in cases where the object the proposition is about (i.e., its subject) undergoes a change.
We arrive at a contradiction in case α + γ = 0. Therefore, we must take α + γ > 0 ; consequently, we have p ( x 2 n + 2, x 2 n + 3 ) < p ( x 2 n + 1, x 2 n + 2 ).
(Of course, if one subscribes to the claim that entailment is explosive, a case for one contradiction is a case for all; but if entailment is paraconsistent, this argument is of no use).
Most paradoxes — but not all — involve contradictions; for such cases, we often use the word "contradiction" as well.
