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Proof step 52 from proof leninv-succ (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 53:
m0 = n0 + m ∧ m ≤ m0, # nσ = ∞, nσ[m0] = nσ[m] + 1, m1 + 1 = n0, m < m0, stutf(nσ), ∀ m0, m, nσ. m1 = m0 - m ∧ # nσ = ∞ ∧ nσ[m0] = nσ[m] + 1 ∧ m < m0 ∧ stutf(nσ) → (∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1) ⊦ ∃ m1. m ≤ m1 ∧ m1 < m0 ∧ nσ[m] = nσ[m1] ∧ nσ[m1 + 1] = nσ[m1] + 1

Rule argument

spec: nat inst: name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [m0, m, n] flags:

Additional information

Used simplifier rules: lels-03: ⊦ m < n → m ≤ n; from specification nat%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: m0, m, n