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Proof step 153 from proof putarray-addarray-override (previous step)

i→n(i0) = n + i→n(i), i ≤ i0, i→n(i) + # vals ≤ # vals0, 0 ≤ i0, i0 < n→i(# vals0), 0 ≤ i, i→n(i) < # vals0 ⊦ (putArray(vals, r0, i, addarray(r0, ty, vals0, st))[r0 -- i0 ']) = (vals + restn(i→n(i) + # vals, vals0))[n], vals0 = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 122:
0 ≤ i ⊦ i→n(i) = ((* m)) ↔ i = n→i(m)
Proof step 152:
0 ≤ i0, i0 = n→i(n + i→n(i)), i ≤ i0, i→n(i) + # vals ≤ # vals0, 0 ≤ i0, i0 < n→i(# vals0), 0 ≤ i, i→n(i) < # vals0 ⊦ (putArray(vals, r0, i, addarray(r0, ty, vals0, st))[r0 -- i0 ']) = (vals + restn(i→n(i) + # vals, vals0))[n], vals0 = []

Rule argument

[speclemmaarg: 0 ≤ i ,i→n(i) = ((* m)) ↔ i = n→i(m) ,[m, i] → [n + i→n(i), i0] ,F ,intnat , ,num-i2n-eq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (n + i→n(i), i0)]]