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Proof step 5 from proof diffone-nodupssize (previous step)

∀ m, a. # x0 = m + 1 ∧ a ∈ x0 → #(x0 \ a ') = m ∨ x0 = [] ∨ dups(x0), #(a0 + x0) = m + 1, a ∈ a0 + x0 ⊦ #((a0 + x0) \ a ') = m, a0 + x0 = [], dups(a0 + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ x0, ∀ m0, a1. # x0 = m0 + 1 ∧ a1 ∈ x0 → #(x0 \ a1 ') = m0 ∨ x0 = [] ⊦ #((a0 ' + x0) \ a ') = # x0, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: diff-firstofapp: ⊦ (x + y) \ x = y \ x;%diff-onenonmember: ⊦ ¬ a ∈ x → x \ a ' = x;%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%e-01: ⊦ a ' + x ≠ []; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%c: ⊦ a + x = a ' + x; from specification list