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Proof step 10 from proof from-append-add (previous step)

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

User applied rule constructor cut

And obtained the following new goals

Proof step 11:
σ = ⌜x0⌝ ⊦ # σ < ⌜n⌝ , (σ to n) + ⌜x⌝ from # x + n = [] ∧ (⌜n0⌝ < #((σ to n) + ⌜x⌝ from # x + n) → ((σ to n) + ⌜x⌝ from # x + n)[n0] = ([])[n0])
Proof step 17:
σ = ⌜nf⌝ ⊦ # σ < ⌜n⌝ , (σ to n) + ⌜x⌝ from # x + n = [] ∧ (⌜n0⌝ < #((σ to n) + ⌜x⌝ from # x + n) → ((σ to n) + ⌜x⌝ from # x + n)[n0] = ([])[n0])

Rule argument

var: σ

Additional information

Used simplifier rules: stream freely generated by ⌜ ⌝, ⌜ ⌝;;