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Proof step 13 from proof getbytearray-subset (previous step)

¬ (r -- (i + n→i(n)) ' ∈ st ∧ st ⊆ st0), n→i(n) < j, n→i(n) < j, n < i→n(j), 0 ≤ i, st ⊆ st0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ st[r -- (i + n→i(n)) '] .byteval = st0[r -- (i + n→i(n)) '] .byteval, j = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
n→i(n) < j, n < i→n(j), 0 ≤ i, st ⊆ st0, 0 < j, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ j = 0, noval .byteval = st0[r -- (i + n→i(n)) '] .byteval, r -- (i + n→i(n)) ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ∅[rk] = noval], [CSIMPSEQ : ⊦ ¬ rk ∈ st → st[rk] = ∅[rk]])]]