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Proof step 56 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 57:
# nσ = ∞, nσ[m0] + 1 = nσ[n], nσ[m0 + 1] = nσ[n], m0 < n, nσ[m0] < nσ[n], stutf(nσ), ∀ m0. n < m0 → nσ[m0] = nσ[n] ⊦ ∃ m1. nσ[m1] = nσ[m0] ∧ (∀ m2. m1 < m2 → nσ[m2] = nσ[m1] + 1)

Rule argument

Additional information

Used simplifier rules: ⊦ ¬ n < n; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m ≤ m; (s)% ⊦ m < n → n ≠ 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat% ⊦ m = n + 1 → ¬ m ≤ n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ ⌜m⌝ < ∞; (s)