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Proof step 29 from proof inv-get-inverse-succ (
previous step
)
∀ n0. nσ[n0] ≠ n, ∃ m. nσ[m] = n, # nσ = ∞, stutf(nσ), ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 30
:
# nσ = ∞, stutf(nσ), ∀ n0. nσ[n0] ≠ nσ[m], ∀ m. ∃ m0. m < m0 ∧ nσ[m] < nσ[m0] ⊦
Rule argument
Additional information