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Proof step 17 from proof getbytearray-subset (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ 0 ≤ j → # getbytearray(r, i, j, st) = i→n(j)
Proof step 16:
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

Rule argument

[emptyarg]

Additional information

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