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Proof step 2 from proof len-04 (previous step)

a ' + x0 = x + b ' + y + c ' + z, # x0 ≤ # x + # z, a ∈ x ∨ a = b ∨ a ∈ y ∨ a = c ∨ a ∈ z ⊦

User applied rule constructor cut

And closed the goal!

Rule argument

var: x

Additional information

Used simplifier rules: Nil: ⊦ [] + x = x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-empty: ⊦ ¬ a ∈ [];%len: ⊦ #(x + y) = # x + # y;%len-05: ⊦ #(a ' + x) = (# x)+1;%c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%len-05: ⊦ #(a ' + x) = (# x)+1;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%len: ⊦ #(x + y) = # x + # y;%list freely generated by [], +;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%lep-02: ⊦ m + n ≤ m + n0 ↔ n ≤ n0; from specification nat%le-num-06: n0 < n ⊦ ¬ m + (* n) ≤ (* n0); from specification nat%size-01: ⊦ # [] = 0; from specification list-data%add-base: ⊦ n + 0 = n; from specification nat-basic2%succ-01: ⊦ m + 1 ≤ n ↔ m < n; from specification nat%l-02: ⊦ ¬ m + n < m; from specification nat-basic2