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Proof step 3 from proof subsize (previous step)

∀ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y)) → x0 = [], # y < #(a + x0), ∀ a0. #oc(a0, a + x0) ≤ #oc(a0, y) ⊦ a + x0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# y ≤ # x0, (∃ y. # y < # x0 ∧ (∀ a. #oc(a, x0) ≤ #oc(a, y))) → x0 = [], ∀ a0. #oc(a0, a ') + #oc(a0, x0) ≤ #oc(a0, y) ⊦

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%s: ⊦ m < n + 1 ↔ ¬ n < m; from specification nat-basic2%len-02: ⊦ # a ' = 0 +1; from specification list%len: ⊦ #(x + y) = # x + # y; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ a + x = a ' + x; from specification list