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Proof step 155 from proof inv-get-inverse-between (previous step)

(∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. ¬(¬ m0 < n0 ∧ nσ[n0] = n)), # nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n ⊦ nσ[m0] = 0, nσ[∞.n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 156:
# nσ = ⌜m0⌝ + ⌜1⌝ , stutf(nσ), n < nσ[m0], ∞.n + m + 1 < max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n, (∀ n0. ∃ n1. n0 < n1 ∧ ¬ m0 < n1 ∧ nσ[n1] = n) ∨ (∀ n0. m0 < n0 ∨ nσ[n0] ≠ n) ⊦ nσ[m0] = 0, nσ[∞.n + m + 1] = n + 1, max∞(λ n0. ¬ m0 < n0 ∧ nσ[n0] = n + 1).n = 0

Proof step 155 from proof inv-get-inverse-between (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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