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Proof step 29 from proof copy-stream-append (previous step)

# buf1 = n2 + n1 + m, ⌜# buf1⌝ = ⌜n1⌝ + ⌜m⌝ + #(⌜buf1⌝ from n1 + m), n4 < n2, n1 < # buf1, n0 + m ≤ # buf0, n1 + m ≤ # buf1 ⊦ buf1[(n4 + m) + n1] = (⌜buf1⌝ from n1 + m)[n4], #(⌜buf0⌝ from n0) < ⌜m⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
⊦ ⌜n⌝ + ⌜m⌝ < # σ0 → (σ0 from n)[m] = σ0[n + m]
Proof step 31:
# buf1 = n2 + n1 + m, ⌜# buf1⌝ = ⌜n1⌝ + ⌜m⌝ + #(⌜buf1⌝ from n1 + m), n1 < # buf1, n4 < n2, ⌜n1 + m⌝ + ⌜n4⌝ < # ⌜buf1⌝ , n1 + m ≤ # buf1, n0 + m ≤ # buf0 ⊦ buf1[(n4 + m) + n1] = ⌜buf1⌝ [(n1 + m) + n4], #(⌜buf0⌝ from n0) < ⌜m⌝

Rule argument

spec: stream, inst: buffer-inf name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ0 → (σ0 from n)[m] = σ0[n + m] subst: [m, n, σ0] → [n4, n1 + m, ⌜buf1⌝ ] flags: <all paths>

Additional information

Used simplifier rules: ⊦ ⌜m + n⌝ = ⌜m⌝ + ⌜n⌝ ; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%add-assoc: ⊦ (n∞ + n∞1) + n∞2 = n∞ + n∞1 + n∞2; from specification nat-inf% ⊦ # ⌜buf⌝ = ⌜# buf⌝ ; (s)% ⊦ ⌜n1⌝ + ⌜n⌝ + ⌜m⌝ < ⌜n0⌝ ↔ n1 + n + m < n0; (s)% ⊦ m = m0 + n0 ∧ n < n0 → m0 + n < m; (s)%add-comm: ⊦ n∞1 + n∞2 = n∞2 + n∞1; from specification nat-inf Used terms: n4, n1 + m, ⌜buf1⌝