Metamath Proof Explorer


Theorem s7eqd

Description: Equality theorem for a length 7 word. (Contributed by Mario Carneiro, 27-Feb-2016)

Ref Expression
Hypotheses s2eqd.1 φ A = N
s2eqd.2 φ B = O
s3eqd.3 φ C = P
s4eqd.4 φ D = Q
s5eqd.5 φ E = R
s6eqd.6 φ F = S
s7eqd.6 φ G = T
Assertion s7eqd φ ⟨“ ABCDEFG ”⟩ = ⟨“ NOPQRST ”⟩

Proof

Step Hyp Ref Expression
1 s2eqd.1 φ A = N
2 s2eqd.2 φ B = O
3 s3eqd.3 φ C = P
4 s4eqd.4 φ D = Q
5 s5eqd.5 φ E = R
6 s6eqd.6 φ F = S
7 s7eqd.6 φ G = T
8 1 2 3 4 5 6 s6eqd φ ⟨“ ABCDEF ”⟩ = ⟨“ NOPQRS ”⟩
9 7 s1eqd φ ⟨“ G ”⟩ = ⟨“ T ”⟩
10 8 9 oveq12d φ ⟨“ ABCDEF ”⟩ ++ ⟨“ G ”⟩ = ⟨“ NOPQRS ”⟩ ++ ⟨“ T ”⟩
11 df-s7 ⟨“ ABCDEFG ”⟩ = ⟨“ ABCDEF ”⟩ ++ ⟨“ G ”⟩
12 df-s7 ⟨“ NOPQRST ”⟩ = ⟨“ NOPQRS ”⟩ ++ ⟨“ T ”⟩
13 10 11 12 3eqtr4g φ ⟨“ ABCDEFG ”⟩ = ⟨“ NOPQRST ”⟩