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Proof step 1 from proof eq-leninv (conclusion)
⊦ ⌜n⌝ < # nσ ∧ stutf(nσ) → (⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n]))
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
⌜n⌝ < # nσ, stutf(nσ) ⊦ ⌜nσ[n]⌝ = leninv(nσ) ↔ # nσ = ∞ ∧ (∀ m. n < m ∧ ⌜m⌝ < # nσ → nσ[m] = nσ[n])
Rule argument
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