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Proof step 6 from proof len-06 (previous step)

# y = # y0, x + y = z + y0 ⊦ # x = # z

User applied rule cut formula

And obtained the following new goals

Proof step 7:
# y = # y0, x + y = z + y0 ⊦ # x = # z, #(x + y) = #(z + y0)

Rule argument

fma: #(x + y) = #(z + y0)

Additional information

Used simplifier rules: len: ⊦ #(x + y) = # x + # y;%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 + y) = #(z + y0)