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Proof step 23 from proof leninv-succ (previous step)

stutf(nσ) ∧ ⌜m0⌝ < # nσ ∧ m3 ≤ nσ[m0] → (∃ m1. m1 ≤ m0 ∧ nσ[m1] = m3), nσ[m0] = m3 + 1, # nσ = ∞, nσ[m] ≤ m3, stutf(nσ), m < m0, ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ ∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1]), m0 = 0, m3 + 1 = nσ[m], ∃ 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 24:
nσ[m0] = nσ[m1] + 1, # nσ = ∞, m1 ≤ m0, nσ[m] ≤ nσ[m1], stutf(nσ), m < m0, ∀ m0. (m0 ≠ 0 → (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1))) → ¬ nσ[m] + 1 < m0 ⊦ m0 = 0, nσ[m0] = nσ[m], ∃ m. nσ[m] = nσ[m1] ∧ ¬(∀ m2. m < m2 → nσ[m2] = nσ[m]), ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%lep: ⊦ m ≤ m + n; from specification nat%less-inf2: ⊦ ⌜m⌝ < ∞; from specification nat-inf