Step
*
2
2
of Lemma
general_length_nth_tl
.....falsecase.....
1. r : ℤ
2. 0 < r
3. ∀[L:Top List]. (||nth_tl(r - 1;L)|| = if r - 1 <z ||L|| then ||L|| - r - 1 else 0 fi ∈ ℤ)
4. L : Top List
5. 0 < r
6. ||L|| ≤ r
⊢ ||nth_tl(r - 1;tl(L))|| = 0 ∈ ℤ
BY
{ xxx(D 4 THEN Reduce 0)xxx }
1
.....wf.....
1. r : ℤ
2. 0 < r
3. ∀[L:Top List]. (||nth_tl(r - 1;L)|| = if r - 1 <z ||L|| then ||L|| - r - 1 else 0 fi ∈ ℤ)
4. 0 < r
5. ||[]|| ≤ r
⊢ r - 1 ∈ ℤ
2
1. r : ℤ
2. 0 < r
3. ∀[L:Top List]. (||nth_tl(r - 1;L)|| = if r - 1 <z ||L|| then ||L|| - r - 1 else 0 fi ∈ ℤ)
4. 0 < r
5. ||[]|| ≤ r
⊢ 0 = 0 ∈ ℤ
3
1. r : ℤ
2. 0 < r
3. ∀[L:Top List]. (||nth_tl(r - 1;L)|| = if r - 1 <z ||L|| then ||L|| - r - 1 else 0 fi ∈ ℤ)
4. u : Top
5. v : Top List
6. 0 < r
7. ||[u / v]|| ≤ r
⊢ ||nth_tl(r - 1;v)|| = 0 ∈ ℤ
Latex:
Latex:
.....falsecase.....
1. r : \mBbbZ{}
2. 0 < r
3. \mforall{}[L:Top List]. (||nth\_tl(r - 1;L)|| = if r - 1 <z ||L|| then ||L|| - r - 1 else 0 fi )
4. L : Top List
5. 0 < r
6. ||L|| \mleq{} r
\mvdash{} ||nth\_tl(r - 1;tl(L))|| = 0
By
Latex:
xxx(D 4 THEN Reduce 0)xxx
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