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Proof step 5 from proof indices-where-in-h (previous step)

m0 +1 ≤ # ar, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ n ∈ ar.indicesWhere(φ, m0 +1) ↔ n < m0 +1 ∧ φ(ar[n])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ n < # ar → ar.indicesWhere(φ, n + 1) = ar.indicesWhere(φ, n) + (φ(ar[n]) ⊃ n + [];[])
Proof step 7:
m0 < # ar, ∀ ar, n, φ. m0 ≤ # ar → (n ∈ ar.indicesWhere(φ, m0) ↔ n < m0 ∧ φ(ar[n])) ⊦ n ∈ ar.indicesWhere(φ, m0) ∨ n ∈ (φ(ar[m0]) ⊃ m0 ';[]) ↔ ¬ m0 < n ∧ φ(ar[n])

Rule argument

Additional information

Used simplifier rules: indiceswhere-rec: ⊦ n < # ar → ar.indicesWhere(φ, n + 1) = ar.indicesWhere(φ, n) + (φ(ar[n]) ⊃ n + [];[]);% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ n ∈ nx + nx0 ↔ n ∈ nx ∨ n ∈ nx0; (s)%a: ⊦ (nx + nx0) + nx1 = nx + nx0 + nx1; from specification list[nat-list-set]% ⊦ nx + [] = nx; (s)% ⊦ n + nx = n ' + nx; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ m +1 = m + 1; (s)