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Proof step 7 from proof restn-firstn (previous step)

x = a + y ∧ x ≠ [], 2 + m0 + n0 + m0 ≤ # x, ∀ n0, x. m0 + n0 + m0 ≤ # x → restn(m0, firstn(n0 + m0, x)) = firstn(n0, restn(m0, x)) ⊦ restn(m0, firstn(n0 + m0, y)) = firstn(n0, restn(m0, y))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%le-num-02: n0 < n ⊦ m + (* n) ≤ m0 + (* n0) ↔ m + (n - n0) ≤ m0; from specification nat%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%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list