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Proof step 4 from proof inl (previous step)

a ∈ a0 + x0, #(a0 + x0) ≤ pos(a, a0 + x0), ∀ a. ¬(a ∈ x0 ∧ # x0 ≤ pos(a, x0)) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a ∈ x0, # x0 ≤ pos(a, x0), ∀ a1. ¬ a1 ∈ x0 ∨ ¬ # x0 ≤ pos(a1, x0) ⊦ a = a0

Rule argument

Additional information

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