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Proof step 8 from proof from-append-add (previous step)

⊦ # σ0 < ⌜n1⌝ , # σ < ⌜n⌝ , #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), ⌜n1⌝ < # σ0 ∧ finp(σ0)

User applied rule constructor cut

And obtained the following new goals

Proof step 9:
σ0 = ⌜x⌝ ⊦ # σ0 < ⌜n1⌝ , # σ < ⌜n⌝ , #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), ⌜n1⌝ < # σ0 ∧ finp(σ0)
Proof step 20:
σ0 = ⌜nf⌝ ⊦ # σ0 < ⌜n1⌝ , # σ < ⌜n⌝ , #((σ to n) + σ0 from n1 + n) = #(σ0 from n1) ∧ (⌜n0⌝ < #((σ to n) + σ0 from n1 + n) → ((σ to n) + σ0 from n1 + n)[n0] = (σ0 from n1)[n0]), ⌜n1⌝ < # σ0 ∧ finp(σ0)

Rule argument

var: σ0

Additional information

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