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Proof step 1 from proof min-stutters (conclusion)
mon<(nσ) ⊦ min(nσ, n + 1) = 1 + min(nσ, n) ∨ min(nσ, n + 1) = min(nσ, n)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
mon<(nσ) ⊦ min(nσ, n + 1) = 1 + min(nσ, n), min(nσ, n + 1) = min(nσ, n)
Rule argument
Additional information