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Proof step 6 from proof subarray-copy (previous step)

m0 + m1 ≤ # ar1 ⊦ # copy(copy(ar0, n0, ar1, n3 + m0, n2), m0, mkarray(m1), 0, m1) = # copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2) ∧ (∀ n. n < # copy(copy(ar0, n0, ar1, n3 + m0, n2), m0, mkarray(m1), 0, m1) → copy(copy(ar0, n0, ar1, n3 + m0, n2), m0, mkarray(m1), 0, m1)[n] = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m0 + m1 ≤ # ar1, n < m1 ⊦ copy(copy(ar0, n0, ar1, n3 + m0, n2), m0, mkarray(m1), 0, m1)[n] = copy(ar0, n0, copy(ar1, m0, mkarray(m1), 0, m1), n3, n2)[n], m1 = 0

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%Size-mkarray: ⊦ # mkarray(n) = n; from specification array