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Proof step 29 from proof inv-get-inverse (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 30:
# 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 ⊦

Rule argument

Additional information

Used simplifier rules: succ: ⊦ m +1 = m + 1; from specification nat-basic2