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Proof step 24 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), ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = ((σ to m + n0) from m)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m]
Proof step 26:
¬ ⌜m + n⌝ < #(σ to n0 + m), #((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m), ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ ((σ to n0 + m) + σ0)[m + n] = ((σ to m + n0) from m)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝
Proof step 30:
⌜m + n⌝ < #(σ to n0 + m), #((σ to n0 + m) + σ0 from m) = #((σ to m + n0) from m) + # σ0, ⌜n⌝ < #((σ to m + n0) from m), ⌜n⌝ < #((σ to m + n0) from m) + # σ0, ⌜n⌝ < ⌜n0⌝ + # σ0 ⊦ (σ to n0 + m)[m + n] = ((σ to m + n0) from m)[n], # σ < ⌜n0⌝ + ⌜m⌝ , # σ < ⌜m⌝ + ⌜n0⌝

Rule argument

spec: inst: name: get-append-ls seq: ⊦ ⌜m⌝ < # σ → (σ + σ1)[m] = σ[m] subst: [m, σ1, σ] → [m + n, σ0, σ to n0 + m] flags: rewrite, current

Additional information

Used terms: m + n, σ0, σ to n0 + m