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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 31:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ nσ0. # nσ0 ≠ ∞ ∨ ¬ stutf(nσ0) ∨ ¬(∀ n1. nσ0[n1] ≠ n0) ∨ ¬(∀ m. ∃ m0. m < m0 ∧ nσ0[m] < nσ0[m0]), ∀ n. nσ[n] ≠ n0 + 1 ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)