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Proof step 8 from proof at-copy (previous step)

m0 < # ar1, ∀ 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 = n0 + m1, # ar1 = 0, copy(ar0, n, ar1, n0, m1)[m0] = (m0 < n0 ∨ n0 + m1 < m0 ⊃ ar1[m0];ar0[m0 + n - n0])

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 9:
m0 < n0 ∨ n0 + m1 < m0, m0 < # ar1, ∀ 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 ⊦ copy(ar0, n, ar1, n0, m1)[m0] = ar1[m0], m0 = n0 + m1, # ar1 = 0
Proof step 10:
¬(m0 < n0 ∨ n0 + m1 < m0), m0 < # ar1, ∀ 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 ⊦ copy(ar0, n, ar1, n0, m1)[m0] = ar0[m0 + n - n0], m0 = n0 + m1, # ar1 = 0

Rule argument

pos: right 3

Proof step 8 from proof at-copy (previous step)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

Additional information