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Proof step 2 from proof len-03 (previous step)

a ' + x = y + b ' + z, # x < # z, a ∈ y ∨ a = b ∨ a ∈ z ⊦ z = []

Heuristic structural induction applied rule structural induction

And closed the goal!

Rule argument

term: y

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 ∈ [];%c: ⊦ a + x = a ' + x;%a: ⊦ (x + y) + z = x + y + z;%c-06: ⊦ a ' + x = b ' + y ↔ a = b ∧ x = y;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x);%len: ⊦ #(x + y) = # x + # y;%len-05: ⊦ #(a ' + x) = (# x)+1;%list freely generated by [], +;;%succ: ⊦ m +1 = m + 1; from specification nat-basic2%l-02: ⊦ ¬ m + n < m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2%ref-1: ⊦ ¬ n < n; from specification nat-basic1