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Proof step 5 from proof inapp (previous step)

a ' + x + b ' = y + c ' + c ' + z, a ∈ y, b ∈ y ∨ b = c ∨ b ∈ z, z ≠ [] → z = y0 + a0 ' ⊦ a = c, a = b, dups(x), b ∈ x, a ∈ x

User applied rule cut formula

And obtained the following new goals

Proof step 6:
a ' + x + b ' = y + c ' + c ' + z, a ∈ y, b ∈ y ∨ b = c ∨ b ∈ z, z ≠ [] → z = y0 + a0 ' ⊦ z = [], a = c, a = b, dups(x), b ∈ x, a ∈ x

Rule argument

fma: z = []

Additional information

Used simplifier rules: app: ⊦ x + [] = x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%app-06: ⊦ x + a ' = y + b ' ↔ x = y ∧ a = b; from specification list%in-14: ⊦ a ' + x = y → a ∈ y; (forward) from specification list%in-15: ⊦ x + a ' = y → a ∈ y; (forward) from specification list%in-11: ⊦ a ∈ x → a ∈ x + y; from specification list%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list Used formulas: z = []