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Proof step 111 from proof inv-get-inverse-succ (previous step)

# nσ = ⌜m⌝ + ⌜1⌝ , n1 < m, nσ[n1] ≤ nσ[m], stutf(nσ), ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ m = 0, nσ[n1 + 1] = nσ[n1] + 1

User applied rule all left

And obtained the following new goals

Proof step 112:
# nσ = ⌜m⌝ + ⌜1⌝ , n1 < m, nσ[n1] ≤ nσ[m], stutf(nσ), ¬ m < n1 + 1 ∧ nσ[n1 + 1] = nσ[n1] → ¬ n1 < n1 + 1, ∀ m0. ¬ m < m0 ∧ nσ[m0] = nσ[n1] → ¬ n1 < m0 ⊦ m = 0, nσ[n1 + 1] = nσ[n1] + 1

Rule argument

pos: left 5 terms: n1 + 1 abort: F discard: F computed: T

Additional information

Used terms: n1 + 1