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Proof step 69 from proof inv-inverts-leq (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 70:
∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[n], # nσ = ⌜m⌝ + ⌜1⌝ , n ≤ m, ∞.n ≤ m, nσ[n] ≤ nσ[m], ∞.n < n, stutf(nσ), ∞.n < m ⊦
Proof step 71:
∀ n0. m < n0 ∨ nσ[n0] ≠ nσ[n], # nσ = ⌜m⌝ + ⌜1⌝ , n ≤ m, ∞.n ≤ m, nσ[n] ≤ nσ[m], ∞.n < n, stutf(nσ), ∞.n < m ⊦

Rule argument

pos: left 8

Proof step 69 from proof inv-inverts-leq (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information