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Proof step 7 from proof del-length (previous step)

# x0 + 1 < #((a0 ' + x0) -l a), ∀ a. # x0 < #(x0 -l a) → x0 -l a = [] ⊦ a = a0, (a0 ' + x0) -l a = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# x0 < #(x0 -l a), ∀ a. # x0 < #(x0 -l a) → x0 -l a = [] ⊦ a = a0, x0 -l a = []

Rule argument

Additional information

Used simplifier rules: del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);%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%l: ⊦ m + n < m + n0 ↔ n < n0; from specification nat-basic2%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list