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Proof step 20 from proof getbytearray-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ bytes = bytes0 ↔ # bytes = # bytes0 ∧ (∀ n. n < # bytes → bytes[n] = bytes0[n])
Proof step 19:
st ⊆ st0, 0 ≤ i, ∀ j0. ¬ j0 < 0 ∧ ¬ i + j ≤ j0 → r -- j0 ' ∈ st ⊦ # getbytearray(r, i, j, st) = # getbytearray(r, i, j, st0) ∧ (∀ n. n < # getbytearray(r, i, j, st) → (getbytearray(r, i, j, st)[n]) = (getbytearray(r, i, j, st0)[n]))

Rule argument

[speclemmaarg: , bytes = bytes0 ↔ # bytes = # bytes0 ∧ (∀ n. n < # bytes → bytes[n] = bytes0[n]) ,[bytes0, bytes] → [getbytearray(r, i, j, st0), getbytearray(r, i, j, st)] ,F ,list-del ,bytelist ,extension ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (getbytearray(r, i, j, st0), getbytearray(r, i, j, st))]]