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Proof step 24 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m0] = m3 + 1, m < m0, stutf(nσ), nσ[m] ≤ m3, stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ m3 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = m3), ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ m3 + 1 = nσ[m], ∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1]), ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 25:
# nσ = ∞, nσ[m0] = nσ[m1] + 1, m < m0, stutf(nσ), m1 ≤ m0, nσ[m] ≤ nσ[m1], ∀ 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), ∃ m. nσ[m] = nσ[m1] ∧ ¬(∀ m2. m < m2 → nσ[m2] = nσ[m])

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ ⌜m⌝ < ∞; (s)