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

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

User applied rule all left

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0; Used terms: 0