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Proof step 2 from proof leq-prefix (
previous step
)
σ1 ≤ σ2 ⊦ σ1 = σ2, ∃ n. ⌜n⌝ < # σ2 ∧ σ2 to n = σ1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
σ1 ≤ σ2 ⊦ σ1 = σ2, ∃ n. ⌜n⌝ < # σ2 ∧ σ2 to n = σ1
Rule argument
Additional information