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Proof step 10 from proof rm-allsize (previous step)

∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 = a ∨ a0 ∈ x0 ↔ a0 ∈ y ⊦ a ∈ x0, # rmdup(a ' + x0) = # rmdup(y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ y. (∀ a. a ∈ x0 ↔ a ∈ y) → # rmdup(x0) = # rmdup(y), ∀ a0. a0 = a ∨ a0 ∈ x0 ↔ a0 ∈ y ⊦ # rmdup(x0) + 1 = # rmdup(y), a ∈ x0

Rule argument

Additional information

Used simplifier rules: rmdup-n: ⊦ ¬ a ∈ x → rmdup(a ' + x) = a + rmdup(x);%succ: ⊦ m +1 = m + 1; from specification nat-basic2%len-05: ⊦ #(a ' + x) = (# x)+1; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list