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Proof step 30 from proof inv-get-inverse (previous step)

# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ nσ0. ¬(∀ n1. nσ0[n1] ≠ n0) ∨ # nσ0 ≠ ∞ ∨ ¬ stutf(nσ0) ∨ ¬(∀ m. ∃ m0. m < m0 ∧ nσ0[m] < nσ0[m0]), ∀ n. nσ[n] ≠ n0 + 1 ⊦

User applied rule all left

And obtained the following new goals

Proof step 31:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ¬(∀ n1. nσ[n1] ≠ n0) ∨ # nσ ≠ ∞ ∨ ¬ stutf(nσ) ∨ ¬(∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0]), ∀ n. nσ[n] ≠ n0 + 1 ⊦

Rule argument

pos: left 4 terms: nσ abort: F discard: T computed: T

Additional information

Used terms: nσ