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Proof step 20 from proof bar-sublist (previous step)

# getbytearray(r, i, n→i(m), st) ≤ m, n→i(n) ≤ i0, 0 ≤ i0 ⊦ sublist(m - # getbytearray(r, i, n→i(m), st), n, getbytearray(r, i + n→i(m), i0, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 19:
m = n0 + # getbytearray(r, i, n→i(m), st), # getbytearray(r, i, n→i(m), st) ≤ m, n→i(n) ≤ i0, 0 ≤ i0 ⊦ sublist(n0, n, getbytearray(r, i + n→i(m), i0, st)) = getbytearray(r, i + n→i(m), n→i(n), st)

Rule argument

[speclemmaarg: n ≤ m ,n0 = m - n ↔ m = n0 + n ,[m, n, n0] → [m, # getbytearray(r, i, n→i(m), st), n0] ,T ,nat , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≤ m ⊦ n0 = m - n ↔ m = n0 + n])], [EXTRATERMS : (m, # getbytearray(r, i, n→i(m), st), n0)]]