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Proof step 32 from proof inv-get-inverse (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n1. nσ[n1] ≠ nσ[n] + 1 ⊦

User applied rule all left

And obtained the following new goals

Proof step 33:
# nσ = ∞, stutf(nσ), ∃ m0. n < m0 ∧ nσ[n] < nσ[m0], ∀ n1. nσ[n1] ≠ nσ[n] + 1 ⊦

Rule argument

pos: left 3 terms: n abort: F discard: T computed: F

Additional information

Used terms: n