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Proof step 20 from proof inv-get-inverse (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 21:
# nσ = ∞, n0 < m, nσ[n0] < nσ[m], stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∃ n2. m < n2 ∧ nσ[n2] = nσ[n0] ⊦ nσ[∞.n] = nσ[n0], n0 = 0

Rule argument

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

Additional information

Used terms: m