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Proof step 6 from proof bytevals-at (previous step)

n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ n = 0, (bytevalues(vals0)[n - 1]) = (val ' + vals0)[n] .byteval

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
n = m + 1, n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ n = 0, (bytevalues(vals0)[m]) = (val ' + vals0)[n] .byteval

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [n, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n, m)]]