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Proof step 19 from proof inv-get-inverse-succ (previous step)

# nσ = ∞, stutf(nσ), ∀ n1. ∃ n2. n1 < n2 ∧ nσ[n2] = nσ[n0], ∀ n. ∃ m. nσ[m] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦ n0 = 0

User applied rule all left

And obtained the following new goals

Proof step 20:
# nσ = ∞, stutf(nσ), ∀ n1. ∃ n2. n1 < n2 ∧ nσ[n2] = nσ[n0], ∀ n. ∃ m. nσ[m] = n, ∃ m0. n0 < m0 ∧ nσ[n0] < nσ[m0] ⊦ n0 = 0

Rule argument

pos: left 5 terms: n0 abort: F discard: T computed: F

Additional information

Used terms: n0