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Proof step 9 from proof bytevals-at (previous step)

∀ n. n < # vals0 → (bytevalues(vals0)[n]) = vals0[n] .byteval ⊦ n < # (val + vals0) → (bytevalues(val + vals0)[n]) = (val + vals0)[n] .byteval

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ bytevalues(val ' + vals) = val .byteval + bytevalues(vals)
Proof step 8:
n ≤ # vals0, ∀ n0. ¬ # vals0 ≤ n0 → (bytevalues(vals0)[n0]) = vals0[n0] .byteval ⊦ (val .byteval ' + bytevalues(vals0))[n] = (val ' + vals0)[n] .byteval

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ # val ' = 0 +1], [CSIMPSEQ : ⊦ # (vals + vals0) = # vals + # vals0], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ by + bytes = by ' + bytes], [CSIMPSEQ : ⊦ val + vals = val ' + vals])]]