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Proof step 3 from proof len-last-rest (previous step)

x + a ' = a0 ' + y, a = a0 ∨ a ∈ y, a0 ∈ x ∨ a0 = a ⊦ # y = # x

User applied rule cut formula

And obtained the following new goals

Proof step 4:
x + a ' = a0 ' + y, a0 ∈ x ∨ a0 = a, a = a0 ∨ a ∈ y ⊦ # y = # x, #(x + a ') = #(a0 ' + y)

Rule argument

fma: #(x + a ') = #(a0 ' + y)

Additional information

Used simplifier rules: len: ⊦ #(x + y) = # x + # y;%len-02: ⊦ # a ' = 0 +1;%a: ⊦ (x + y) + z = x + y + z;% ⊦ m + n = m + n0 ↔ n = n0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used formulas: #(x + a ') = #(a0 ' + y)