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Proof step 2 from proof at-copy (previous step)

m0 < # ar1 ⊦ copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]), # ar1 = 0

User applied rule structural induction

And obtained the following new goals

Proof step 3:
m0 < # ar1 ⊦ copy(ar0, n, ar1, n0, 0)[m0] = (m0 < n0 ∨ n0 + 0 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]), # ar1 = 0
Proof step 5:
∀ m0, ar1, ar0, n, n0. m0 < # ar1 → copy(ar0, n, ar1, n0, m1)[m0] = (m0 < n0 ∨ n0 + m1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]) ∨ # ar1 = 0, m0 < # ar1 ⊦ copy(ar0, n, ar1, n0, m1 +1)[m0] = (m0 < n0 ∨ n0 + m1 +1 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]), # ar1 = 0

Rule argument

indvar: m

Additional information

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