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Proof step 1 from proof leq-prefix (conclusion)
⊦ σ1 ≤ σ2 → σ1 = σ2 ∨ (∃ n. ⌜n⌝ < # σ2 ∧ σ2 to n = σ1)
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 2
:
σ1 ≤ σ2 ⊦ σ1 = σ2, ∃ n. ⌜n⌝ < # σ2 ∧ σ2 to n = σ1
Rule argument
Additional information