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deduce (something) from (something)

To infer information from something. Oh, I deduced from her disinterested tone that she wouldn't be joining us today. A: "Did you know he wasn't coming?" B: "I deduced that when I saw you pull up alone."
See also: deduce
Farlex Dictionary of Idioms. © 2022 Farlex, Inc, all rights reserved.

deduce something from something

to infer or conclude something from a set of facts. Can I deduce a bit of anger from your remarks? I deduce nothing from everything I have heard today.
See also: deduce
McGraw-Hill Dictionary of American Idioms and Phrasal Verbs. © 2002 by The McGraw-Hill Companies, Inc.
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References in periodicals archive ?
As [x.sup.n] > 0, we deduce by assumption with m = 0 that f (s, [x.sup.n.sub.s]) [greater than or equal to] -[gamma][x.sup.n.sub.s] for all s [member of] [0,[T.sub.0]] hence
By induction, we then deduce that [mathematical expression not reproducible] for all t [member of] [0,[T.sub.0]].
By induction, we deduce that [[parallel]x - y[parallel].sub.[0,t]] [less than or equal to] 2C[k.sup.n][t.sup.n]/n!
Then, we also deduce that the sequence [([x.sup.m]).sub.m[greater than or equal to]1] uniformly converges to some function [z.sup.n] on [[t.sub.n], [t.sub.n] + [T.sup.*.sub.0]] such that
Therefore, by definition of [T.sup.*], we deduce that [t.sub.n] + [T.sup.*.sub.0] [less than or equal to] [T.sup.*] for all n.
remarking via the hypotheses that 10[x.sub.n,1] = n + 71, clearly 10[x.sub.n,1] < n + 71, now using the previous inequality and (4), then it becomes trivial to deduce that statement Z([x.sub.n,1], [gamma](n), [gamma]'(n)) is true, otherwise, Z(xni1,7(n),7'(n)) is false, and using (4), then we clearly deduce that it is false that 10xn1 < n + 71, therefore, 10xn1 > n + 71.
In this case, using the definition of Y([x.sub.n,1], [gamma](n), [gamma]'(n)) (see Definitions 2.1), then we immediately deduce that statement Y([x.sub.n,1], [gamma](n), [gamma]'(n)) is of the form
Indeed, remarking by (5) that [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII], and recalling that we are in the case where Z([x.sub.n,1], [gamma](n), [gamma]'(n)) is true, then, using the previous, it becomes trivial to deduce that Z([x.sub.n,1], [gamma](n), [gamma]'(n)) and Y([x.sub.n,1], [gamma](n), [gamma]'(n)), are simultaneously true.
Otherwise, using the definition of statement Z([x.sub.n,1], [gamma](n), [gamma]'(n)) (see Definitions 2.1), then we immediately deduce that statement Z([x.sub.n,1], [gamma](n), [gamma]'(n)) is of the form
Here, using only the immediate part of the generalized Fermat induction, simple definitions, elementary arithmetic congruences, elementary complex analysis, elementary arithmetic calculus, reasoning by reduction to absurd and properties (2.4) and (2.3) of Remark 2, we prove a Theorem which implies the Mersenne primes conjecture; moreover, from our Theorem, we immediately deduce that the Mersenne primes conjecture that we solved, was only an elementary consequence of the Goldbach conjecture.
For that, let n + 11; recalling that in particular n is of type 37, clearly (by using the definition of type 37), n [equivalent to] 0 mod(37) and using the previous congruence, we immediately deduce that
That being so, if [m.sub.n,1] <n + 11, then, using congruences (R.1.5) and (R.1.4), it becomes trivial to deduce that the previous inequality immediately implies that
Now consider the quantity R[Z(n)] + I[Z(n)]; then, using (R.1.7), it becomes immediate to deduce that
Cullen deduces what he can from prefatory references suggesting associations with the Pierrepont family of Dorchester and with James Howell, author of Dodona's Grove; other scholars will no doubt discover additional facts in due course.
He insists on a dark power as the cause and iconography of men in black: "Alone or in ranks, the man in black is the agent of a serious power; and of a power claimed over women and the feminine." Accordingly, he deduces gravely, "We live now in the aftertow of the black wave's latest rise and breaking....