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Proof step 24 from proof leninv-inf (previous step)

∀ n. ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ ∃ n1. n0 < n1 ∧ (∃ m. nσ[m] = n1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n1 - 1))

User applied rule exists right

And obtained the following new goals

Proof step 25:
∀ n. ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], n ≠ 0 → nσ[m] = n - 1 ∧ ¬(m < m0 → nσ[m0] = n - 1) ⊦ n0 < n0 +1 ∧ (∃ m. nσ[m] = n0 +1 - 1 ∧ ¬(∀ m0. m < m0 → nσ[m0] = n0 +1 - 1))

Rule argument

pos: right 1 terms: n0 +1 abort: F discard: T computed: T

Additional information

Used terms: n0 +1