Sentence examples for proof argues from inspiring English sources

Exact(1)

A third sort of a priori proof argues from facts about the mere possibility of something (or of some collection of things) to the existence of a "ground of possibility" that somehow explains them.

Similar(59)

During a heated debate in the judge's robing room, Mr. Quattrone's lawyer, John W. Keker, objected to the judge's removing the references to the burden of proof, arguing that "there is a tremendous amount of favorable instruction to the defense that has been stripped away out of this Allen charge," according to a transcript.

We complete the proof arguing as in the proof of Theorem 4.1 □.

Bolzano [1817] expressed this attitude with regard to the intermediate value theorem for the real numbers (IVT) before giving a purely analytic proof, arguing that spatial thinking could not be used to help justify the IVT.

That's a sign that storage may be relatively recession-proof, argues Sun's Fowler.

Kant, having rejected the cosmological, ontological, and design proofs, argued in the Critique of Practical Reason (1788) that the existence of God, though not directly provable, is a necessary postulate of the moral life.

Proof Argue by contradiction.

Proof Arguing indirectly, assume that for some sequence { u k } ⊂ E ˜ with ∥ u k ∥ → ∞, and there is M 2 > 0 such that Φ ( u k ) ≥ − M 2 for all k ∈ N. Set v k = u k / ∥ u k ∥, then ∥ v k ∥ = 1.

Theorem 4.1 There is no nontrivial traveling-wave solution u ∈ C ( [ 0, ∞ ) ; H 3 ) ∩ C 1 ( [ 0, ∞ ) ; H 2 ) for equation (4.1) with γ = − 2 ω α 2. Proof Arguing by contradiction, assume that w ∈ H 3 and u ( t, x ) = w ( x − c t ), c ≠ 0 is a strong solution of (4.1).

Lemma 2.7 Φ is coercive on ℳ, that is, Φ ( x ) → ∞ as ∥ x ∥ → ∞, x ∈ M. Proof Arguing by contradiction, suppose there exists a sequence { x j } ⊂ M such that ∥ x j ∥ → ∞ and Φ ( x j ) ≤ d for some d ∈ [ c, ∞ ). Let y j : = x j / ∥ x j ∥. Then y j ⇀ y and y j → y a.e. t ∈ [ 0, T ] as j → ∞ after passing to a subsequence.

Proof Arguing along the lines of the proof of Theorem 3, we see that ‖ h 1 / 2 (1 − A h ) ϕ h ‖ L 2 ≲ ‖ ϕ h − Ψ h ‖ H ˜ − 1 / 2 for all Ψ h ∈ S 1 (T h ).

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