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Proof step 8 from proof mkjavavalues-restn (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ 1 + # bytes0, ∀ n0. ¬ # bytes0 < n0 → restn(n0, mkjavavalues(bytes0)) = mkjavavalues(restn(n0, bytes0)) ⊦ restn(n, byteval(by) ' + mkjavavalues(bytes0)) = mkjavavalues(restn(n - 1, bytes0)), n = 0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < n ↔ n ≠ 0], [CSIMPSEQ : ⊦ (bytes + bytes0) + bytes1 = bytes + bytes0 + bytes1], [CSIMPSEQ : ⊦ (by ' + bytes) .rest = bytes], [CSIMPSEQ : ⊦ n -1 = n - 1])]]