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Proof step 14 from proof leninv-succ (previous step)

# nσ = ∞, stutf(nσ), ∀ m. (∃ m0. nσ[m0] = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)) → ¬ n + 1 < m, ∃ m. nσ[m] = n ∧ ¬(∀ m0. m < m0 → nσ[m0] = nσ[m]) ⊦ ∃ m. nσ[m] = n ∧ (∀ m0. m < m0 → nσ[m0] = nσ[m] + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# nσ = ∞, m < m0, stutf(nσ), ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ nσ[m0] = nσ[m], ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic