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Proof step 12 from proof swap-to-from-app (previous step)

#((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0 ⊦ ((σ to n0 + m) + σ0 from m)[n] = (((σ to m + n0) from m) + σ0)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 13:
⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m]
Proof step 14:
¬ ⌜m⌝ + ⌜n⌝ < #((σ to n0 + m) + σ0), #((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0 ⊦ ((σ to n0 + m) + σ0 from m)[n] = (((σ to m + n0) from m) + σ0)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝
Proof step 20:
⌜m⌝ + ⌜n⌝ < #((σ to n0 + m) + σ0), #((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m) + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = (((σ to m + n0) from m) + σ0)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

spec: inst: name: get-from seq: ⊦ ⌜n⌝ + ⌜m⌝ < # σ → (σ from n)[m] = σ[n + m] subst: [m, n, σ] → [n, m, (σ to n0 + m) + σ0] flags: rewrite, current

Additional information

Used terms: n, m, (σ to n0 + m) + σ0