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Proof step 4 from proof len (previous step)

a0 < a, ∀ a. # ins≤(a, x0) = # x0 + 1 ⊦ # ins≤(a, a0 ' + x0) = (1 + # x0) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 < a, ∀ a. # ins≤(a, x0) = # x0 + 1 ⊦ # ins≤(a, x0) = # x0 + 1

Rule argument

Additional information

Used simplifier rules: ins-n: ⊦ b < a → ins≤(a, b ' + x) = b + ins≤(a, x);%n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)