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Proof step 3 from proof len-01 (previous step)

∀ a. # x0 < pos(a, x0) → pos(a, x0) = 0, #(a0 + x0) < pos(a, a0 + x0) ⊦ pos(a, a0 + x0) = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x0 + 1 < pos(a, a0 ' + x0), ∀ a. # x0 < pos(a, x0) → pos(a, x0) = 0 ⊦ pos(a, a0 ' + x0) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list