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Proof step 10 from proof bytevalues-eq (previous step)

by ' + bytevalues(vals0) = bytevalues(vals1), by ∈ bytevalues(vals1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0), bytevalues(vals0) = bytevalues(vals1 .rest) ∧ (val ∈ vals0 ∨ val ∈ vals1 .rest → is_bytevalue(val)) → vals0 = vals1 .rest ⊦ byteval(by) ' + vals0 = vals1

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
⊦ bytevalues([]) = []
Proof step 9:
vals1 = val1 + vals ∧ vals1 ≠ [], by ' + bytevalues(vals0) = bytevalues(vals1), by ∈ bytevalues(vals1), ∀ val0. val0 ∈ vals0 ∨ val0 ∈ vals1 → is_bytevalue(val0), bytevalues(vals0) = bytevalues(vals) ∧ (val ∈ vals0 ∨ val ∈ vals → is_bytevalue(val)) → vals0 = vals ⊦ byteval(by) ' + vals0 = vals1

Rule argument

[speclemmaarg: vals ≠ [] ,val = vals .first ∧ vals0 = vals .rest ↔ vals = val + vals0 ,[vals, val, vals0] → [vals1, val1, vals] ,T ,list-data ,javavalues ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ bytes + bytes0 ≠ [] ↔ bytes = [] ∧ bytes0 = []], [CSIMPSEQ : ⊦ by ' ≠ []], [CSIMPSEQ : ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1], [CSIMPELIM : vals ≠ [] ⊦ val = vals .first ∧ vals0 = vals .rest ↔ vals = val + vals0])], [EXTRATERMS : (vals1, val1, vals)]]