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Proof step 136 from proof putarray-addarray-override (previous step)

0 ≤ i, i→n(i) < # vals0, # vals + i→n(i) ≤ # vals0, i→n(i) + n0 + # vals < # vals0, 0 ≤ (n→i(n0) + n→i(# vals)) + i, (# vals + i→n(i)) + n0 < # vals0, 0 ≤ i + n→i(# vals) ⊦ vals0 = [], vals0[i→n(i) + n0 + # vals] = (restn(# vals + i→n(i), vals0)[n0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 133:
⊦ (m + n0) + n1 = m + n0 + n1
Proof step 135:
0 ≤ i, i→n(i) < # vals0, # vals + i→n(i) ≤ # vals0, # vals + i→n(i) + n0 < # vals0, 0 ≤ (n→i(n0) + n→i(# vals)) + i, # vals + i→n(i) + n0 < # vals0, 0 ≤ i + n→i(# vals) ⊦ vals0 = [], vals0[# vals + i→n(i) + n0] = (restn(# vals + i→n(i), vals0)[n0])

Rule argument

[speclemmaarg: ,(m + n0) + n1 = m + n0 + n1 ,[n1, n0, m] → [n0, i→n(i), # vals] ,F ,nat-basic2 , ,a ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (n0, i→n(i), # vals)]]