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Proof step 26 from proof copy-stream-append (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 27:
# buf1 = n2 + n1 + m, ⌜# buf1⌝ = ⌜n1⌝ + ⌜m⌝ + #(⌜buf1⌝ from n1 + m), n1 < # buf1, n3 < n2 + m, m ≤ n3, n1 + m ≤ # buf1, n0 + m ≤ # buf0 ⊦ buf1[n3 + n1] = (⌜buf1⌝ from n1 + m)[n3 - m], n2 = 0, #(⌜buf0⌝ from n0) < ⌜m⌝

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < m + n ↔ n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)% ⊦ ⌜m⌝ .n = m; (s)