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Proof step 2 from proof sorted-lt (previous step)

m0 < m, m1 < m, sorted(zbrar, 0, m) ⊦ zbrar[m0].key < zbrar[m1].key ↔ m0 < m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m0 < m, m1 < m, sorted(zbrar, 0, m) ⊦ zbrar[m0].key < zbrar[m1].key ↔ m0 < m1

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic