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Proof step 23 from proof get-sublist (previous step)

n→i(m) + n→i(n) ≤ j ⊦ sublist(m, n, getarray(r, i, j, st)) = getarray(r, i + n→i(m), n→i(n), st)

User applied rule structural induction

And obtained the following new goals

Proof step 17:
⊦ n→i(0) + n→i(n) ≤ j → sublist(0, n, getarray(r, i, j, st)) = getarray(r, i + n→i(0), n→i(n), st)
Proof step 22:
∀ n, j, r, i, st. n→i(m0) + n→i(n) ≤ j → sublist(m0, n, getarray(r, i, j, st)) = getarray(r, i + n→i(m0), n→i(n), st) ⊦ n→i(m0 +1) + n→i(n) ≤ j → sublist(m0 +1, n, getarray(r, i, j, st)) = getarray(r, i + n→i(m0 +1), n→i(n), st)

Rule argument

[VARWITHFMAVARSARG : m, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPGEN : nat generated by 0, +1;])]]