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Proof step 16 from proof leninv-zero (previous step)

# nσ = ∞, stutf(nσ), ∀ m. (∃ m0. nσ[m0] = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)) → m = 0 ⊦ nσ[n] = 0, nσ[0] = 0 ∧ ¬(∀ m0. 0 < m0 → nσ[m0] = 0), ∃ m. nσ[m] = 0 ∧ ¬(∀ m0. m < m0 → nσ[m0] = 0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
# nσ = ∞, stutf(nσ), ∀ m. (∃ m0. nσ[m0] = m - 1 ∧ ¬(∀ m1. m0 < m1 → nσ[m1] = m - 1)) → m = 0, ∀ m. m ≠ 0 → nσ[m] = 0 ⊦ nσ[n] = 0, ∃ m. nσ[m] = 0 ∧ ¬(∀ m0. m < m0 → nσ[m0] = 0)

Rule argument

Additional information

Used simplifier rules: stutf-zero: ⊦ stutf(nσ) → nσ[0] = 0;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1