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Proof step 47 from proof inv-inverts-leq (previous step)

# nσ = ∞, nσ[n0] = nσ[n], ∞.n < n, nσ[n] ≤ nσ[m], stutf(nσ), m + 1 < n0, ∀ 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 48:
# nσ = ∞, nσ[n0] = nσ[n], ∞.n < n, nσ[n] ≤ nσ[m], stutf(nσ), m + 1 < n0, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1, ∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n] ⊦

Rule argument

pos: left 8 terms: m + 1 abort: F discard: T computed: F

Additional information

Used terms: m + 1