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Proof step 18 from proof putar-getar (previous step)

i < i + 1 ∨ (i + 1) + i1 ≤ i, n→i(# vals0) = i1 + 1, 0 ≤ i1, ∀ r0, i0, st0. getarray(r0, i0, i1 + 1, putArray(vals0, r0, i0, st0)) = vals0 ⊦ vals0 = [], getarray(r, i + 1, i1, putArray(vals0, r, i + 1, st)) + vals0[i→n(i1)] ' = vals0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
n→i(# vals0) = i1 + 1, 0 ≤ i1, ∀ r0, i0, st0. getarray(r0, i0, i1 + 1, putArray(vals0, r0, i0, st0)) = vals0 ⊦ vals0 = [], getarray(r, i + 1, i1, putArray(vals0, r, i + 1, st)) + vals0[i→n(i1)] ' = vals0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i ≤ i], [CSIMPSEQ : ⊦ i < j + 1 ↔ i ≤ j])]]