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Proof step 28 from proof inv-get-inverse (previous step)

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

User applied rule structural induction

And obtained the following new goals

Proof step 29:
# nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∀ n0. nσ[n0] ≠ 0 ⊦
Proof step 30:
# 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])) ⊦

Rule argument

indvar: n

Additional information

Used simplifier rules: nat freely generated by 0, +1;;