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Proof step 3 from proof put-del1-in (previous step)

m < #(a0 + x0), ∀ m, a. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a ∈ (a0 + x0)[m, a] -1l a ↔ a ∈ a0 + x0, (a0 + x0)[m] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
m ≤ # x0, ∀ m, a. m < # x0 → (a ∈ x0[m, a] -1l a ↔ a ∈ x0) ∨ x0[m] = a ⊦ a ∈ (a0 ' + x0)[m, a] -1l a ↔ a ∈ a0 ' + x0, (a0 ' + x0)[m] = a

Rule argument

Additional information

Used simplifier rules: ⊦ m < n ↔ ¬ n ≤ m; (ss)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m < n + 1 ↔ ¬ n < m; (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ # a ' = 0 +1; (s)% ⊦ #(x + y) = # x + # y; (s)% ⊦ a + x = a ' + x; (s)