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Proof step 2 from proof stutf-alltaken (
previous step
)
⌜m⌝ < # nσ, stutf(nσ), n ≤ nσ[m] ⊦ ∃ m0. m0 ≤ m ∧ nσ[m0] = n
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
⌜m⌝ < # nσ, stutf(nσ), n ≤ nσ[m] ⊦ ∃ m0. m0 ≤ m ∧ nσ[m0] = n
Rule argument
Additional information