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Proof step 35 from proof eq-leninv (previous step)

nσ[m2] + 1 = nσ[n], # nσ = ∞, nσ[n] < nσ[m0], n < m0, stutf(nσ), m2 < m0 → nσ[m0] = nσ[n], ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = 0, m0 = 0, nσ[m0] = nσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 36:
# nσ = ∞, nσ[m2] + 1 = nσ[n], n < m2, m0 ≤ m2, stutf(nσ), n < m0, nσ[n] < nσ[m0], ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = nσ[n], m0 = 0, nσ[m0] = 0, m2 = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat