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Proof step 46 from proof inv-inverts-leq (previous step)

∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], # nσ = ∞, stutf(nσ), nσ[n] ≤ nσ[m], ∞.n < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 47:
# nσ = ∞, nσ[n0] = nσ[n], ∞.n < n, nσ[n] ≤ nσ[m], stutf(nσ), m + 1 < n0, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1