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Proof step 19 from proof leninv-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 20:
# 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)

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [nσ[m0], m3]

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: nσ[m0], m3