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Proof step 23 from proof inv-inverts-leq (previous step)

# nσ = ∞, ∞.n + 1 < m, n < m, nσ[n] < nσ[m], ∞.n < n, ∞.n < m, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0], ∃ n1. m < n1 ∧ nσ[n1] = nσ[n] ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
# nσ = ∞, nσ[n0] = nσ[n], ∞.n < n0, ∞.n + 1 < m, n < m, ∞.n < n, ∞.n < m, m < n0, (∞.n + 1) + 1 < n0, n + 1 < n0, n < n0, nσ[n] < nσ[m], ∞.n + 1 < n0, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦

Rule argument

Additional information

Used simplifier rules: ⊦ m +1 = m + 1; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic%trans: ⊦ n < n0 ∧ n0 < n1 → n < n1; (forward) from specification nat-basic%lf: ⊦ m < n ∧ n < n0 → m +1 < n0; (forward) from specification nat-basic