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Proof step 5 from proof append-mod (previous step)

∀ n0, x0, a, x1. n0 < # x0 → x0 = [] ∨ ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ # x2 + # x0 = 0, n0 < # x0 ⊦ x0 = [], ((a0 + x2) + x0)[n0 + #(a0 + x2), a] = (a0 + x2) + x1 ↔ x0[n0, a] = x1, #(a0 + x2) + # x0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → x0 = [] ∨ ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ # x2 + # x0 = 0 ⊦ x0 = [], (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1

Rule argument

Additional information

Used simplifier rules: put-succ: ⊦ (b ' + x)[n + 1, a] = b + x[n, a];%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%app-08: ⊦ a ' + x = a ' + y ↔ x = y; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list