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Proof step 53 from proof leninv-succ (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 54:
# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m] + 1, 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) ⊦ ∃ m0. m ≤ m0 ∧ ¬ m1 + m < m0 ∧ nσ[m] = nσ[m0] ∧ nσ[m0 + 1] = nσ[m0] + 1

Rule argument

Additional information

Used simplifier rules: l-02: ⊦ ¬ m + n < m; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%lels-03: ⊦ m < n → m ≤ n; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2