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Proof step 9 from proof getbytearray-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
j ≤ n→i(n), 0 < j, st ⊆ st0, 0 ≤ i, n < i→n(j), ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ j = 0, (getbytearray(r, i, j, st)[n]) = (getbytearray(r, i, j, st0)[n]), n = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : 0 ≤ i ⊦ ((* i)) < n→i(n) ↔ i→n(i) < n], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 < i → 0 < j], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i])]]