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Proof step 2 from proof delete-one-nls (previous step)

∀ a. ¬ # x0 < #(x0 -1l a) ⊦ ¬ #(a0 + x0) < #((a0 + x0) -1l a)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
# x0 + 1 < #((a0 ' + x0) -1l a), ∀ a. ¬ # x0 < #(x0 -1l a) ⊦ (a0 ' + x0) -1l a = []

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%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