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Proof step 14 from proof from-to-rest (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# buf = n + m ⊦ ⌜m⌝ = #(⌜buf⌝ from n), #(⌜buf⌝ from n) < ⌜m⌝

Rule argument

Additional information

Used simplifier rules: ⊦ # σ0 = ⌜n⌝ → σ0 to n = σ0; (s)% ⊦ ⌜m⌝ < ⌜n⌝ ↔ m < n; (s)