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Proof step 6 from proof del1 (previous step)

#oc(a, a0 + x0) = m + 1, a ∈ a0 + x0, ∀ a, m. #oc(a, x0) = m + 1 ∧ a ∈ x0 → #oc(a, x0 -1l a) = m ⊦ #oc(a, (a0 + x0) -1l a) = m

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#oc(a, x0) = m + 1, a ∈ x0, ∀ a, m. #oc(a, x0) = m + 1 ∧ a ∈ x0 → #oc(a, x0 -1l a) = m ⊦ #oc(a, x0 -1l a) = m, a = a0

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%one: ⊦ #oc(a, a ') = 1;%one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%oc-n: ⊦ a ≠ b → #oc(a, b ' + x) = #oc(a, x);% ⊦ a ≠ b → (b ' + x) -1l a = b ' + (x -1l a); (s)% ⊦ n + 0 = n; (s)% ⊦ (a ' + x) -1l a = x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)