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Proof step 17 from proof copy-stream-append (previous step)

# buf1 = n2 + n1 + m, n = n3 + n1, ⌜# buf1⌝ = ⌜n1⌝ + #((⌜buf0⌝ from n0) to m) + #(⌜buf1⌝ from n1 + m), n1 < # buf1, n < # buf1, n1 ≤ n, n1 + m ≤ # buf1, n0 + m ≤ # buf0 ⊦ n2 + m = 0, (n1 + m ≤ n ⊃ buf1[n];buf0[n3 + n0]) = (((⌜buf0⌝ from n0) to m) + (⌜buf1⌝ from n1 + m))[n3]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
# buf1 = n2 + n1 + m, ⌜# buf1⌝ = ⌜n1⌝ + #((⌜buf0⌝ from n0) to m) + #(⌜buf1⌝ from n1 + m), n3 < n2 + m, n1 < # buf1, n0 + m ≤ # buf0, n1 + m ≤ # buf1 ⊦ (m ≤ n3 ⊃ buf1[n3 + n1];buf0[n3 + n0]) = (((⌜buf0⌝ from n0) to m) + (⌜buf1⌝ from n1 + m))[n3]

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m ≤ m + n; (s)%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)