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Proof step 8 from proof subarray-longer-eq-eq (previous step)

# copy(ar0, 0, mkarray(n + m), 0, n + m) = # copy(ar1, 0, mkarray(n + m), 0, n + m), n0 < n, ∀ n0. n0 < # copy(ar0, 0, mkarray(n + m), 0, n + m) → copy(ar0, 0, mkarray(n + m), 0, n + m)[n0] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n0] ⊦ copy(ar0, 0, mkarray(n), 0, n)[n0] = copy(ar1, 0, mkarray(n), 0, n)[n0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n0 < n, ∀ n1. n1 < n + m → copy(ar0, 0, mkarray(n + m), 0, n + m)[n1] = copy(ar1, 0, mkarray(n + m), 0, n + m)[n1] ⊦ copy(ar0, 0, mkarray(n), 0, n)[n0] = copy(ar1, 0, mkarray(n), 0, n)[n0]

Rule argument

Additional information

Used simplifier rules: size-copy: ⊦ # copy(ar, m, ar0, m0, n) = # ar0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ # mkarray(n) = n; (s)