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Proof step 5 from proof from-append-tofrom (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝
Proof step 7:
¬ ¬ #(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , #((σ from n) to n1) = ⌜n1⌝ ∧ (⌜n0⌝ < ⌜n⌝ + #((σ from n) to n1) → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0])
Proof step 12:
¬ #(σ from n) < ⌜n1⌝ ⊦ # σ < ⌜n⌝ + ⌜n1⌝ , ⌜n1⌝ = ⌜n1⌝ ∧ (⌜n0⌝ < ⌜n⌝ + ⌜n1⌝ → ((σ to n) + ((σ from n) to n1))[n0] = (σ to n + n1)[n0])

Rule argument

spec: inst: name: size-to seq: ¬ # σ < ⌜n⌝ ⊦ #(σ to n) = ⌜n⌝ subst: [n, σ] → [n1, σ from n] flags: rewrite, current

Additional information

Used terms: n1, σ from n