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Proof step 3 from proof rmtoobigpos (previous step)

∀ n. # x0 ≤ n → x0 -1l n = x0, #(a + x0) ≤ n ⊦ (a + x0) -1l n = a + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x0 < n, ∀ n. # x0 ≤ n → x0 -1l n = x0 ⊦ (a ' + x0) -1l n = a ' + x0, n = 0

Rule argument

Additional information

Used simplifier rules: 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%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list