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Proof step 4 from proof len (previous step)

a0 < a, ∀ a. # ins≤(a, x0) = # x0 + 1 ⊦ # ins≤(a, a0 ' + x0) = 2 + # x0

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);%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-num-01: n < n0 ⊦ m + (* n) = m0 + (* n0) ↔ m = m0 + (n0 - n); from specification nat%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list