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Proof step 12 from proof move-initvals (previous step)

i→n(i) = # vals1, 0 ≤ i, ∀ val, st, ty, r, i, vals0. # vals0 < # vals1 + 1 → (true → (0 ≤ i ∧ i→n(i) < # vals0 → (addarray(r, ty, vals0, st)[r -- i ', val]) = addarray(r, ty, vals0[i→n(i), val], st))) ⊦ (addarray(r, ty, vals1, st)[r -- i ', val][r -- n→i(# vals1) ', val0][r -- _length, intval(n→i(# vals1) + 1)]) = addarray(r, ty, vals1 + val ', st), i = n→i(# vals1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
0 ≤ i ⊦ i→n(i) = m ↔ i = n→i(m)

Rule argument

[speclemmaarg: 0 ≤ i ,i→n(i) = ((* m)) ↔ i = n→i(m) ,[m, i] → [# vals1, i] ,F ,intnat , ,num-i2n-eq ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (# vals1, i)]]