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Proof step 1 from proof at-shl (conclusion)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: shl: ⊦ shl(zbrar, n, m) = copy(zbrar, n + 1, zbrar, n, m - n - 1);%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n - m = n; (s)%⊦ m0 < # zbrar → copy(zbrar0, n, zbrar, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ zbrar[m0];zbrar0[m0 + n - n0]); (s)