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Proof step 46 from proof inv-inverts-leq (previous step)

# nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n] ⊦

User applied rule all left

And obtained the following new goals

Proof step 47:
# nσ = ∞, ∞.n < n, stutf(nσ), nσ[n] ≤ nσ[m], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

pos: left 6 terms: m + 1 abort: F discard: F computed: F

Additional information

Used terms: m + 1