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Proof step 41 from proof stutf-maxinf-is-fin (previous step)

# nσ = ∞, nσ[n2] = nσ[n], stutf(nσ), n2 ≤ n0, ∀ n1. ¬ n1 < n2 → nσ[n1] = nσ[n] ⊦ n0 < n0 +1 ∧ nσ[n0 +1] = nσ[n]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 42:
# nσ = ∞, nσ[n2] = nσ[n], stutf(nσ), n2 ≤ n0, ∀ n1. ¬ n1 < n2 → nσ[n1] = nσ[n] ⊦ nσ[n0 + 1] = nσ[n]

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2