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Proof step 45 from proof inv-inverts-leq (previous step)

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

User applied rule all left

And obtained the following new goals

Proof step 46:
∃ n1. m + 1 < n1 ∧ nσ[n1] = nσ[n], ∀ n0. ∃ n1. n0 < n1 ∧ nσ[n1] = nσ[n], # nσ = ∞, stutf(nσ), nσ[n] ≤ nσ[m], ∞.n < n, ∀ m0. m < m0 → nσ[m0] = nσ[m] + 1 ⊦

Rule argument

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

Additional information

Used terms: m + 1