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Proof step 6 from proof from-to-rest (previous step)

# buf = n + m ⊦ #((⌜buf⌝ from n) to m) = #(⌜buf⌝ from n) ∧ (⌜n0⌝ < #((⌜buf⌝ from n) to m) → ((⌜buf⌝ from n) to m)[n0] = (⌜buf⌝ from n)[n0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝
Proof step 8:
# buf = n + m, #(⌜buf⌝ from n) < ⌜m⌝ ⊦ #((⌜buf⌝ from n) to m) = #(⌜buf⌝ from n) ∧ (⌜n0⌝ < #((⌜buf⌝ from n) to m) → ((⌜buf⌝ from n) to m)[n0] = (⌜buf⌝ from n)[n0])
Proof step 14:
# buf = n + m ⊦ #(⌜buf⌝ from n) < ⌜m⌝ , ⌜m⌝ = #(⌜buf⌝ from n) ∧ (⌜n0⌝ < ⌜m⌝ → ((⌜buf⌝ from n) to m)[n0] = (⌜buf⌝ from n)[n0])

Rule argument

spec: stream, inst: buffer-inf name: size-to seq: ¬ # σ0 < ⌜n⌝ ⊦ #(σ0 to n) = ⌜n⌝ subst: [n, σ0] → [m, ⌜buf⌝ from n] flags: <all paths>

Additional information

Used terms: m, ⌜buf⌝ from n