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Proof step 24 from proof range-copy-outer (previous step)

n1 < m1, (n3 + n0) + n1 < # ar0, (n + n3) + m1 ≤ # ar, n2 + m1 ≤ # ar2, n3 + m1 ≤ m0, (n3 + n0) + m1 ≤ # ar0, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, ∀ n4. n4 < m1 → ar2[n2 + n4] = ar[(n + n3) + n4] ⊦ ar2[n2 + n1] = ((n3 + n0) + n1 < n0 ∨ n0 + m0 ≤ (n3 + n0) + n1 ⊃ ar0[(n3 + n0) + n1];ar[((n3 + n0) + n1) + n - n0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
n1 < m1, (n3 + n0) + n1 < # ar0, (n + n3) + m1 ≤ # ar, n2 + m1 ≤ # ar2, n3 + m1 ≤ m0, (n3 + n0) + m1 ≤ # ar0, n0 + m0 ≤ # ar0, n + m0 ≤ # ar, ∀ n4. n4 < m1 → ar2[n2 + n4] = ar[(n + n3) + n4] ⊦ ar2[n2 + n1] = ar[n3 + n1 + n]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m + n - m = n; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ m + n < m; (s)