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Proof step 36 from proof eq-leninv (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 37:
# nσ = ∞, nσ[m2] + 1 = nσ[n], nσ[n] < nσ[m0], n < m0, n < m2, stutf(nσ), m0 ≤ m2, ∀ m1. m2 < m1 → nσ[m1] = nσ[n] ⊦ nσ[m0] = nσ[n]

Rule argument

Additional information

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