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Proof step 6 from proof mkjavavalues-restn (previous step)

n = m + 1, n ≤ 1 + # bytes0, ∀ n0. ¬ # bytes0 < n0 → restn(n0, mkjavavalues(bytes0)) = mkjavavalues(restn(n0, bytes0)) ⊦ restn(n, byteval(by) ' + mkjavavalues(bytes0)) = mkjavavalues(restn(m, bytes0)), n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
m ≤ # bytes0, ∀ n0. ¬ # bytes0 < n0 → restn(n0, mkjavavalues(bytes0)) = mkjavavalues(restn(n0, bytes0)) ⊦ restn(m, mkjavavalues(bytes0)) = mkjavavalues(restn(m, bytes0))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ m + n ≤ m + n0 ↔ n ≤ n0], [CSIMPSEQ : ⊦ m + 1 ≠ 0], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ restn(n + 1, val ' + vals) = restn(n, vals)])]]