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

m1 < m, sorted(zbrar, n, m) ⊦ zbrar[n].key < zbrar[n + m1].key ↔ 0 ≠ m1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m1 < m, sorted(zbrar, n, m) ⊦ 0 = m1, zbrar[n].key < zbrar[n + m1].key

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ ¬ key < key; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)