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Proof step 3 from proof inl (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
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;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < 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%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%c: ⊦ a + x = a ' + x; from specification list