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Proof step 48 from proof gcd-partial (previous step)

n = N, m = M ⊦ [: N, M | (N' = N'' ∧ M' = M''), true, true, GCD(; N, M)] (gcd(n, m) = M ∧ N = gcd(n, m))

User applied rule tl call right

And obtained the following new goals

Proof step 47:
n = N, m = M ⊦ [: N, M | (N' = N'' ∧ M' = M''), true, true, GCD_loop(; N, M) ∥ GCD_loop(; M, N)] (gcd(n, m) = M ∧ N = gcd(n, m))

Rule argument

pos: right 1

Additional information

[Concretesimprules((Csimpdecl(GCD(; N, M) {GCD_loop(; N, M) ∥ GCD_loop(; M, N)})))]