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Proof step 10 from proof copy-subarray2 (previous step)

n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ m1 = 0, copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = (n + m0 < m0 + n1 ∨ (m0 + n1) + n2 ≤ n + m0 ⊃ ar1[n + m0];ar0[(n + m0) + n0 - m0 + n1])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n + m0 < # ar1, m0 + m1 ≤ # ar1, n < m1 ⊦ m1 = 0, copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n1, n2)[n] = (n < n1 ∨ n1 + n2 ≤ n ⊃ ar1[n + m0];ar0[n + n0 - n1]), # ar1 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%cancel-01: ⊦ m + n - n + n1 = m - n1; from specification nat%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2