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Proof step 18 from proof inv-get-inverse-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 19:
# nσ = ∞, stutf(nσ), ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = n, ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n. ∃ m. nσ[m] = n, ∃ n1. 0 < n1 ∧ nσ[n1] = n ⊦

Rule argument

pos: left 5 terms: 0 abort: F discard: F computed: F

Additional information

Used terms: 0