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Proof step 24 from proof leninv-zero (previous step)

# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = m - 1) → m = 0 ⊦

User applied rule all left

And obtained the following new goals

Proof step 25:
# nσ = ⌜n0⌝ + ⌜1⌝ , stutf(nσ), (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = nσ[n0] + 1 - 1) → nσ[n0] + 1 = 0, ∀ m. (∃ m0. ¬ n0 < m0 ∧ nσ[m0] = m - 1) → m = 0 ⊦

Rule argument

pos: left 3 terms: nσ[n0] + 1 abort: F discard: F computed: F

Additional information

Used terms: nσ[n0] + 1