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Proof step 1 from proof sublist-append-02-01 (conclusion)
⊦ # x ≤ m ∧ m ≤ # x + # y → sublist(0, m, x + y) = x + sublist(0, m - # x, y)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
# x ≤ m, m ≤ # x + # y ⊦ sublist(0, m, x + y) = x + sublist(0, m - # x, y)
Rule argument
Additional information