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Proof step 51 from proof putarray-addarray-01 (previous step)

i→n(i1) = i→n(i) + n, 0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0, i1 < i + n→i(# vals) + i0, 0 ≤ i1, 0 < i + n→i(# vals) + i0, 0 < n→i(# vals) + i0, i ≤ i1, n→i(# vals) + i ≤ i1 ⊦ initial_value(ty) = (vals + mklist(initial_value(ty), i→n(i0)))[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 39:
0 ≤ i ⊦ i→n(i) = ((* m)) ↔ i = n→i(m)
Proof step 50:
0 ≤ i1, i1 = n→i(i→n(i) + n), 0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0, i1 < i + n→i(# vals) + i0, 0 ≤ i1, 0 < i + n→i(# vals) + i0, 0 < n→i(# vals) + i0, i ≤ i1, n→i(# vals) + i ≤ i1 ⊦ initial_value(ty) = (vals + mklist(initial_value(ty), i→n(i0)))[n]

Rule argument

[speclemmaarg: 0 ≤ i ,i→n(i) = ((* m)) ↔ i = n→i(m) ,[m, i] → [i→n(i) + n, i1] ,F ,intnat , ,num-i2n-eq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (i→n(i) + n, i1)]]