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Proof step 15 from proof inv-get-inverse (previous step)

∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = n

User applied rule all left

And obtained the following new goals

Proof step 16:
∃ n1. 0 < n1 ∧ nσ[n1] = n, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ nσ[∞.n] = n

Rule argument

pos: left 1 terms: 0 abort: F discard: F computed: F

Additional information

Used terms: 0