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Proof step 1 from proof sublist-num-01 (conclusion)
n ≠ 0 ⊦ sublist(0, (* n), a ' + x) = a ' + sublist(0, n -1, x)
Special step normalize
And obtained the following new goals
Proof step 2
:
n ≠ 0 ⊦ sublist(0, n, a ' + x) = a ' + sublist(0, n -1, x)