[Imprint][Data Privacy Statement][E-Mail]

View long theorem info. View the proof. Back to main index. View the specification. View theorem base summary, View theorem base listing, the used properties by spec, the dependency list, the symbol table,

Proof step 17 from proof put-sw-in (previous step)

m = m1 + 1, m0 < # x0, m ≤ # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ x0 = [], m = 0, a ∈ (a0 ' + x0)[m, x0[m0]][m0 + 1, x0[m1]] ↔ a = a0 ∨ a ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 18:
m0 < # x0, m1 < # x0, ∀ m, n, a. m < # x0 ∧ n < # x0 → (a ∈ x0[m, x0[n]][n, x0[m]] ↔ a ∈ x0) ⊦ x0 = [], a ∈ x0[m1, x0[m0]][m0, x0[m1]] ↔ a ∈ x0, a = a0

Rule argument

Additional information

Used simplifier rules: put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2