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Proof step 2 from proof sublist-lenrec (previous step)

m + n < # x ⊦ sublist(m, n + 1, x) = sublist(m, n, x) + x[m + n]

User applied rule structural induction

And obtained the following new goals

Proof step 3:
0 + n < # x ⊦ sublist(0, n + 1, x) = sublist(0, n, x) + x[0 + n]
Proof step 5:
m0 +1 + n < # x, ∀ n, x. m0 + n < # x → sublist(m0, n + 1, x) = sublist(m0, n, x) + x[m0 + n] ⊦ sublist(m0 +1, n + 1, x) = sublist(m0 +1, n, x) + x[m0 +1 + n]

Rule argument

indvar: m

Additional information

Used simplifier rules: nat freely generated by 0, +1;;