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Proof step 14 from proof subarray-elim (previous step)

n + m ≤ # ar, ∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar1 = m ⊦ ar1 = subarray(ar, n, m)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 16:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 17:
n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m)
Proof step 18:
⊦ ar1 = subarray(ar2 + ar1, # ar2, # ar1)

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%subarray-elim-right: n + m ≤ # ar0 ⊦ subarray(ar0 + ar1, n, m) = subarray(ar0, n, m);% ⊦ m ≤ m + n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ ¬ n < n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)