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Proof step 18 from proof emptystream-cancel (previous step)

# σ = ⌜n⌝ , ∀ n0. ⌜n0⌝ < ⌜n⌝ + # σ⊥ → (σ + σ⊥)[n0] = zero, # σ⊥ = ∞ ∨ σ = [] ⊦ ⌜n⌝ = #(σ⊥ to n) ∧ (n0 < n → σ[n0] = (σ⊥ to n)[n0])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 19:
⊦ σ⊥ = ⌜λ n. zero⌝
Proof step 20:
# σ = ⌜n⌝ , ∀ n0. ⌜n0⌝ < ⌜n⌝ + # ⌜λ n. zero⌝ → (σ + ⌜λ n. zero⌝ )[n0] = zero, # ⌜λ n. zero⌝ = ∞ ∨ σ = [] ⊦ ⌜n⌝ = #(⌜λ n. zero⌝ to n) ∧ (n0 < n → σ[n0] = (⌜λ n. zero⌝ to n)[n0])

Rule argument

spec: <toplevel> name: emptystream seq: ⊦ σ⊥ = ⌜λ n. zero⌝ subst: flags: <all paths>

Additional information

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