Step
*
of Lemma
rprod_functionality
No Annotations
∀[n,m:ℤ]. ∀[x,y:{n..m + 1-} ⟶ ℝ]. rprod(n;m;k.x[k]) = rprod(n;m;k.y[k]) supposing x[k] = y[k] for k ∈ [n,m]
BY
{ (Auto THEN Assert ∀d:ℕ. (((n + d) ≤ m)
⇒ (rprod(n;n + d;k.x[k]) = rprod(n;n + d;k.y[k])))⋅) }
1
.....assertion.....
1. n : ℤ
2. m : ℤ
3. x : {n..m + 1-} ⟶ ℝ
4. y : {n..m + 1-} ⟶ ℝ
5. x[k] = y[k] for k ∈ [n,m]
⊢ ∀d:ℕ. (((n + d) ≤ m)
⇒ (rprod(n;n + d;k.x[k]) = rprod(n;n + d;k.y[k])))
2
1. n : ℤ
2. m : ℤ
3. x : {n..m + 1-} ⟶ ℝ
4. y : {n..m + 1-} ⟶ ℝ
5. x[k] = y[k] for k ∈ [n,m]
6. ∀d:ℕ. (((n + d) ≤ m)
⇒ (rprod(n;n + d;k.x[k]) = rprod(n;n + d;k.y[k])))
⊢ rprod(n;m;k.x[k]) = rprod(n;m;k.y[k])
Latex:
Latex:
No Annotations
\mforall{}[n,m:\mBbbZ{}]. \mforall{}[x,y:\{n..m + 1\msupminus{}\} {}\mrightarrow{} \mBbbR{}].
rprod(n;m;k.x[k]) = rprod(n;m;k.y[k]) supposing x[k] = y[k] for k \mmember{} [n,m]
By
Latex:
(Auto THEN Assert \mforall{}d:\mBbbN{}. (((n + d) \mleq{} m) {}\mRightarrow{} (rprod(n;n + d;k.x[k]) = rprod(n;n + d;k.y[k])))\mcdot{})
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