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Proof step 12 from proof prefixn-split-sublist (previous step)

# y = n0 + n1 ⊦ y = firstn(n0, y) + sublist(n0, n1, y)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ n + m = # x → sublist(n, m, x) = restn(n, x)
Proof step 14:
n0 + n1 = # y ⊦ y = firstn(n0, y) + restn(n0, y)

Rule argument

spec: list-del name: sublist-restn-02 seq: ⊦ n + m = # x → sublist(n, m, x) = restn(n, x) subst: [x, m, n] → [y, n1, n0] flags: <all paths>

Additional information

Used terms: y, n1, n0