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Proof step 1 from proof sublist-two-num (conclusion)
m1 ≤ # x, n + n1 ≤ m1 ⊦ sublist((* n), (* n1), sublist(0, (* m1), x)) = sublist(n, n1, x)
Special step normalize
And obtained the following new goals
Proof step 2
:
m1 ≤ # x, n + n1 ≤ m1 ⊦ sublist(n, n1, sublist(0, m1, x)) = sublist(n, n1, x)