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Proof step 3 from proof getbar-appeq (conclusion)

⊦ 0 ≤ j → ( getbytearray(r, i, j, st) + bytes = getbytearray(r0, i0, j, st) + bytes0 ↔ getbytearray(r, i, j, st) = getbytearray(r0, i0, j, st) ∧ bytes = bytes0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ # bytes = # bytes0 → (bytes + bytes1 = bytes0 + bytes2 ↔ bytes = bytes0 ∧ bytes1 = bytes2)
Proof step 2:
⊦ 0 ≤ j → # getbytearray(r, i, j, st) = i→n(j)

Rule argument

[speclemmaarg: , # bytes = # bytes0 → ( bytes + bytes1 = bytes0 + bytes2 ↔ bytes = bytes0 ∧ bytes1 = bytes2) ,[bytes2, bytes0, bytes1, bytes] → [bytes0, getbytearray(r0, i0, j, st), bytes, getbytearray(r, i, j, st)] ,F ,list ,bytelist ,len-01 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # bytes ≠ # bytes0 → bytes ≠ bytes0], [CSIMPSEQ : ⊦ i→n(i) = i→n(j) ↔ abs(i) = abs(j)])], [EXTRATERMS : (bytes0, getbytearray(r0, i0, j, st), bytes, getbytearray(r, i, j, st))]]