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Proof step 28 from proof inv-get-inverse-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 29:
∀ n0. nσ[n0] ≠ n, ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Rule argument

pos: left 2 terms: n abort: F discard: T computed: T

Additional information

Used terms: n