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Proof step 8 from proof putbytearray-addarray-override (previous step)

# vals1 = 1 + # bytes0, ∀ vals, vals2, r0, st0. # vals2 = # bytes0 → putByteArray(bytes0, r0, n→i(# vals), addarray(r0, byte_type, vals + vals2, st0)) = addarray(r0, byte_type, vals + mkjavavalues(bytes0), st0) ⊦ vals1 = [], putByteArray(bytes0, r, n→i(# vals0) + 1, addarray(r, byte_type, vals0 + byteval(by) ' + vals1 .rest, st)) = addarray(r, byte_type, vals0 + byteval(by) ' + mkjavavalues(bytes0), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 7:
vals1 = val + vals3, # vals1 = 1 + # bytes0, ∀ vals, vals2, r0, st0. # vals2 = # bytes0 → putByteArray(bytes0, r0, n→i(# vals), addarray(r0, byte_type, vals + vals2, st0)) = addarray(r0, byte_type, vals + mkjavavalues(bytes0), st0) ⊦ vals1 = [], putByteArray(bytes0, r, n→i(# vals0) + 1, addarray(r, byte_type, vals0 + byteval(by) ' + vals3, st)) = addarray(r, byte_type, vals0 + byteval(by) ' + mkjavavalues(bytes0), st)

Rule argument

[speclemmaarg: vals ≠ [] ,val = vals .first ∧ vals0 = vals .rest ↔ vals = val + vals0 ,[vals, val, vals0] → [vals1, val, vals3] ,T ,list-data ,javavalues ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : vals ≠ [] ⊦ val = vals .first ∧ vals0 = vals .rest ↔ vals = val + vals0])], [EXTRATERMS : (vals1, val, vals3)]]