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Proof step 3 from proof range-eq-shift-right (previous step)

⊦ (# ar0 + m) + m1 ≤ #(ar0 + ar1) ∧ m0 + m1 ≤ # ar ∧ (∀ n. n < m1 → (ar0 + ar1)[(# ar0 + m) + n] = ar[m0 + n]) ↔ range=(ar1, m, ar, m0, m1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ #(ar0 + ar1) = # ar0 + # ar1
Proof step 5:
⊦ (ar + ar0) + ar1 = ar + ar0 + ar1
Proof step 6:
n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n]
Proof step 7:
⊦ m + m1 ≤ # ar1 ∧ m0 + m1 ≤ # ar ∧ (∀ n. n < m1 → ar1[m + n] = ar[m0 + n]) ↔ range=(ar1, m, ar, m0, m1)

Rule argument

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-a: ⊦ (ar + ar0) + ar1 = ar + ar0 + ar1;%append-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];% ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)%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)