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Proof step 19 from proof get-get (previous step)

1 < j, n→i(n) < j, ∀ j0, n, r, i, st. i→n(j0) < i→n(j) → (n→i(n) < j0 → (getarray(r, i, j0, st)[n]) = st[r -- (i + n→i(n)) ']) ⊦ (getarray(r, i, j - 1, st) + st[r -- (i + j - 1) '])[n] = st[r -- (i + n→i(n)) ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
1 < j, n→i(n) < j, ∀ j0, n, r, i, st. i→n(j0) < i→n(j) → (n→i(n) < j0 → (getarray(r, i, j0, st)[n]) = st[r -- (i + n→i(n)) ']) ⊦ (getarray(r, i, j - 1, st) + st[r -- (i + j - 1) '] ')[n] = st[r -- (i + n→i(n)) ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ vals + val = vals + val '])]]