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Proof step 2 from proof at-shr (previous step)

m0 < # zbrar, n ≤ m ⊦ (¬ n < m0 ∨ m < m0 ⊃ zbrar[m0];zbrar[m0 - 1]) = (m0 ≤ n ∨ m < m0 ⊃ zbrar[m0];zbrar[m0 - 1])

User applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ m < n → (n ≤ m ↔ false); (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)