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Proof step 7 from proof bytevals-at (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ n = 0, (bytevalues(vals0)[n - 1]) = (val ' + vals0)[n] .byteval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ n -1 = n - 1])]]