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Proof step 29 from proof inv-ls-succ (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 30:
# nσ = ∞, stutf(nσ), ∀ n. nσ[n] ≠ 1, ∃ m0. 0 < m0 ∧ nσ[0] < nσ[m0], ∀ n0, nσ. n0 < 0 → # nσ = ∞ ∧ stutf(nσ) ∧ (∀ n1. nσ[n1] ≠ n0 + 1) ∧ (∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]) → false ⊦

Rule argument

pos: left 3 terms: 0 abort: F discard: T computed: F

Additional information

Used terms: 0