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Proof step 2 from proof one-1init-02 (conclusion)

⊦ (addarray(r, ty, val ' + val0 ' + vals, st)[r -- 1 ']) = val0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
0 ≤ i, i < n→i(# vals) ⊦ (addarray(r, ty, vals, st)[r -- ((* i)) ']) = vals[i→n(i)]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (val ' + vals)[0] = val], [CSIMPSEQ : n ≠ 0 ⊦ (val ' + vals)[(* n)] = vals[n -1]], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ ((* j)) < i0 + ((* j0)) ↔ j - j0 < i0], [CSIMPSEQ : ⊦ (~1) < i ↔ 0 ≤ i], [CSIMPSEQ : ⊦ 0 ≤ n→i(n)], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i])]]