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Proof step 3 from proof unequal (previous step)

∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1), # x = #(a + y0) ⊦ x = a + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a + y0 = z + b ' + z1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
# x = # y0 + 1, ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x = a ' + y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ x = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Rule argument

Additional information

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