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Proof step 2 from proof mininv-inverts (
previous step
)
min(nσ, n + 1) = 1 + min(nσ, n), mon<(nσ) ⊦ nσ[min(nσ, n)] = n
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
min(nσ, n + 1) = 1 + min(nσ, n), mon<(nσ) ⊦ nσ[min(nσ, n)] = n
Rule argument
Additional information