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Proof step 6 from proof append-mod (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n0 < # x0, ∀ n0, x0, a, x1. n0 < # x0 → ((x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1) ∨ x0 = [] ⊦ (x2 + x0)[n0 + # x2, a] = x2 + x1 ↔ x0[n0, a] = x1, x0 = []

Rule argument

Additional information

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