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Proof step 54 from proof leninv-succ (previous step)

# 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

User applied rule all left

And obtained the following new goals

Proof step 55:
# nσ = ∞, nσ[(m1 + 1) + m] = nσ[m] + 1, stutf(nσ), m1 = m1 + m - m ∧ # nσ = ∞ ∧ nσ[m1 + m] = nσ[m] + 1 ∧ m < m1 + m ∧ stutf(nσ) → (∃ m2. m ≤ m2 ∧ m2 < m1 + m ∧ nσ[m] = nσ[m2] ∧ nσ[m2 + 1] = nσ[m2] + 1), ∀ 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

pos: left 4 terms: m1 + m, m, nσ abort: F discard: F computed: F

Additional information

Used terms: m1 + m, m, nσ