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Proof step 21 from proof inv-get-inverse-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 22:
# nσ = ∞, stutf(nσ), n0 < m, nσ[n0] < nσ[m], ∃ n2. m < n2 ∧ nσ[n2] = nσ[n0], ∀ n. ∃ m. nσ[m] = n ⊦ n0 = 0, m = 0, nσ[m] = 0

Rule argument

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

Additional information

Used terms: m