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Proof step 8 from proof del-length (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
# x0 < #(x0 -l a), ∀ a. # x0 < #(x0 -l a) → x0 -l a = [] ⊦ x0 -l a = [], a = a0

Rule argument

Additional information

Used simplifier rules: del-n: ⊦ a ≠ b → (b ' + x) -l a = b ' + (x -l a);% ⊦ # x = 0 ↔ x = []; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a ' + x ≠ []; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list