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Proof step 20 from proof leninv-succ (previous step)

# nσ = ∞, nσ[m0] = m3 + 1, nσ[m] < nσ[m0], m < m0, stutf(nσ), (∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m3)) → ¬ nσ[m] + 1 < nσ[m0], ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ nσ[m0] = nσ[m], nσ[m0] = 0, ∃ 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 21:
# nσ = ∞, nσ[m0] = m3 + 1, m < m0, stutf(nσ), nσ[m] ≤ m3, (∃ m1. nσ[m1] = m3 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = nσ[m1])) → ¬ nσ[m] < m3, ∀ m0. (∃ m1. nσ[m1] = m0 - 1 ∧ ¬(∀ m2. m1 < m2 → nσ[m2] = m0 - 1)) → ¬ nσ[m] + 1 < m0 ⊦ m3 + 1 = 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 < m ↔ false); (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)