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Proof step 12 from proof unequal (previous step)

# x0 = # y0, # x0 = # y0 → x0 = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x0 = z + a ' + z0 ∧ y0 = z + b ' + z1), ∀ x. # x = # y0 → x = y0 ∨ (∃ z, a, b, z0, z1. a ≠ b ∧ x = z + a ' + z0 ∧ y0 = z + b ' + z1) ⊦ x0 = y0, ∃ z, a0, b, z0, z1. a0 ≠ b ∧ a ' + x0 = z + a0 ' + z0 ∧ a ' + y0 = z + b ' + z1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
# a0 ' + # z0 = # b ' + # z1, ∀ x. # x = # z + # b ' + # z1 → x = z + b ' + z1 ∨ (∃ z2, a, b0, z0, z3. a ≠ b0 ∧ x = z2 + a ' + z0 ∧ z + b ' + z1 = z2 + b0 ' + z3) ⊦ a0 ' + z0 = b ' + z1, a0 = b, ∃ z2, a1, b0, z4, z3. a1 ≠ b0 ∧ a ' + z + a0 ' + z0 = z2 + a1 ' + z4 ∧ a ' + z + b ' + z1 = z2 + b0 ' + z3

Rule argument

Additional information

Used simplifier rules: len: ⊦ #(x + y) = # x + # y;%a: ⊦ (x + y) + z = x + y + z;%app-01: ⊦ x + y = x + z ↔ y = z;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)