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Proof step 3 from proof at-copy (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m0 < # ar1 ⊦ # ar1 = 0, ar1[m0] = (m0 < n0 ∨ n0 ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0])

Rule argument

Additional information

Used simplifier rules: Copy-zero: ⊦ copy(ar, m, ar0, m0, 0) = ar0;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2