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Proof step 27 from proof leninv-succ (previous step)

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, m1 < m0 → nσ[m0] = nσ[m1] ⊦ m0 = 0, nσ[m0] = nσ[m], ∃ m0. nσ[m0] = nσ[m] ∧ (∀ m1. m0 < m1 → nσ[m1] = nσ[m0] + 1)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 8

Additional information

Used simplifier rules: eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%anti: ⊦ m ≤ n → (n ≤ m ↔ m = n); from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1