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Proof step 8 from proof bytevals-at (previous step)

n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ (val .byteval ' + bytevalues(vals0))[n] = (val ' + vals0)[n] .byteval

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
n ≠ 0 ⊦ (by ' + bytes)[(* n)] = bytes[n -1]
Proof step 7:
n ≠ 0, n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ (bytevalues(vals0)[n -1]) = (val ' + vals0)[n] .byteval

Rule argument

[speclemmaarg: n ≠ 0 ,(by ' + bytes)[(* n)] = bytes[n -1] ,[n, bytes, by] → [n, bytevalues(vals0), val .byteval] ,F ,list-del ,bytelist ,get-numrec ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (by ' + bytes)[0] = by], [CSIMPSEQ : ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1], [CSIMPSEQ : ⊦ (val ' + vals)[0] = val], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1])], [EXTRATERMS : (n, bytevalues(vals0), val .byteval)]]