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Proof step 6 from proof len-01 (previous step)

1 + # x0 < pos(a, a0 ' + x0), ∀ a. ¬ # x0 < pos(a, x0) ⊦ a0 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
# x0 < pos(a, x0), ∀ a. ¬ # x0 < pos(a, x0) ⊦ a0 = a

Rule argument

Additional information

Used simplifier rules: pos-n: ⊦ a ≠ b → pos(a, b ' + x) = pos(a, x) + 1;% ⊦ m < n → n ≠ 0; (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: ⊦ (x + y) + z = x + y + z; from specification list