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Proof step 103 from proof inv-get-inverse-succ (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 104:
∀ n0. ∃ n1. n0 < n1 ∧ ¬ m < n1 ∧ nσ[n1] = nσ[m0], # nσ = ⌜m⌝ + ⌜1⌝ , m0 ≤ m, ∞.n < m, nσ[m0] ≤ nσ[m], stutf(nσ) ⊦ m = 0, nσ[∞.n + 1] = nσ[m0] + 1
Proof step 105:
∀ n0. m < n0 ∨ nσ[n0] ≠ nσ[m0], # nσ = ⌜m⌝ + ⌜1⌝ , m0 ≤ m, ∞.n < m, nσ[m0] ≤ nσ[m], stutf(nσ) ⊦ m = 0, nσ[∞.n + 1] = nσ[m0] + 1

Rule argument

pos: left 6

Proof step 103 from proof inv-get-inverse-succ (previous step)

User applied rule case distinction

And obtained the following new goals

Rule argument

Additional information